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THE ELECTRIC CIRCUIT 



WORKS BY THE SAME AUTHOR 



Published by McGRAW-HllX BOOK COMPANY 

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THE 

ELECTRIC CIRCUIT 



BY 

V. KARAPETOFF 



SECOND EDITION 

REWRITTEN, ENLARGED AND ENTIRELY RESET 



McGRAW-HILL BOOK COMPANY 

239 WEST 39TH STREET, NEW YORK 

6 BOUVERIE STREET, LONDON, E.C. 

1912 



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Copyright, 1912, by the 
McGRAW-HILL BOOK COMPANY 



Copyright, 1910, by V. KARAPETOFF 



Stanbope jpress 

F. H.GILSON COMPANY 
BOSTON, U.S.A. 



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€fl.A316198 I 



PREFACE TO THE FIRST EDITION 



This pamphlet, together with the companion pamphlet en- 
titled The Magnetic Circuit, is intended to give a student in 
electrical engineering the theoretical elements necessary for cal- 
culation of the performance of dynamo-electric machinery and 
of transmission lines. The advanced student must be taught to 
treat every electric machine as a particular combination of electric 
and magnetic circuits, and to base its performance upon the 
fundamental theoretical relations rather than upon a separate 
" theory " established for each kind of machinery, as is often 
done. 

The first chapter is devoted to a review of the direct-current 
circuit, the next four chapters treat of sine-wave alternating- 
current circuits, and the last two chapters give the fundamental 
properties of the electrostatic circuit. All the important results 
and methods are illustrated by numerical problems of which there 
are over one hundred in the text. The pamphlet is not intended 
for a beginner, but for a student who has had an elementary de- 
scriptive course in electrical engineering and some simple labora- 
tory experiments. 

The treatment is made as far as possible uniform, so that the 
student sees analogous relations in the direct-current circuit, in 
the alternating-current circuit, in the electrostatic circuit and 
finally in the magnetic circuit. All matter of purely historical or 
academic interest, not bearing directly upon the theory of electric 
machinery, has been left out. An ambitious student will find a 
more exhaustive treatment in the works mentioned at the end of 
the pamphlet. 

The electrostatic circuit is treated in accordance with the 
modern conception of elastic displacement of electricity in di- 
electrics. No use has been made of the action of electric charges 
at a distance, or of the electrostatic system of units. The volt- 
ampere-ohm system of units is used for electrostatic calculations, 



VI PREFACE TO THE FIRST EDITION 

in accordance with Professor Giorgi's ideas (see. a paper by Pro- 
fessor Ascoli in Vol. I of the Transactions of the International 
Electrical Congress, St. Louis, 1904). Those familiar with 
Oliver Heaviside's writings will notice his influence upon the 
author, particularly in Arts. 22 and 23,* where an attempt is 
made at a rational electrostatic nomenclature. 

Many thanks are due to the author's friend and colleague, 
Mr. John F. H. Douglas, instructor in electrical engineering in 
Sibley College, who read the manuscript and the proofs, checked 
the answers to the problems and made many excellent suggestions 
for the text. 

Cornell University, Ithaca, N. Y. 
August, 1910. 



PREFACE TO THE SECOND EDITION 



The first edition of this book was issued as a pamphlet of 
some 85 pages which the author used for two years in his classes 
to supplement some other texts. In its present edition, the book 
is made independent of these texts, so that its size had to be 
more than doubled. The book has been practically rewritten, and 
completely reset in type. All the cuts are new. The topics are 
treated somewhat more in detail, and a large number of practical 
problems are provided. The new topics added are : the resistance 
of conductors of variable cross-section, the electrical relations in 
polyphase systems, performance characteristics of the trans- 
mission line, transformer and induction motor and the permittance 
(electrostatic capacity) of transmission lines. 

In the treatment of alternating currents by means of complex 
quantities, particular attention is paid to the trigonometric form 
E (cos 6 -f- j sin 0) of the expression for a vector. In fact, the 
transmission line, the transformer, and the induction motor to 
some extent, are treated in this trigonometric form. The author 
trusts that the reader will find this somewhat novel treatment 
more convenient in numerical applications than the usual form 
e + je'. 

* Chapter 14 in the second edition. 



PREFACE TO THE SECOND EDITION Vll 

Since the appearance of the first edition, the author has been 
encouraged by some of his colleagues in his treatment of the 
electrostatic circuit in the ampere-ohm system of units, a treat- 
ment which involves the use of permittances in farads and 
elastances in darafs. He has extended this treatment to the calcu- 
lation of capacity of cables and transmission lines. The students 
grasp this mode of presentation much more readily than the old- 
fashioned way, based upon the law of inverse squares and elec- 
tric charges acting at a distance. The purpose of the present 
treatment is to impress them with the idea of a continuous action 
in the medium itself and with the role of the dielectric. 

Mr. F. R. Keller of the electrical department of Columbia 
University has read and corrected the manuscript and the proofs 
of the second edition, and checked the answers to the new prob- 
lems. The author wishes to express sincere appreciation of his 
painstaking, faithful and competent work. The author is also 
indebted to Mr. John F. H. Douglas for critically reading the 
galley proof of the second edition. 

Cornell University, Ithaca, N. Y. 
May, 1912. 



CONTENTS 



PAGE 

Preface v 

Suggestions to Teachers xi 

Chapter I. Fundamental Electrical Relations in Direct-cur- 
rent Circuits 1 

The volt, the ampere, the ohm, and the mho. Temperature Co- 
efficient. Resistances and conductances in series and in parallel. 
Electric power. 
Chapter II. Fundamental Electrical Relations in Direct-cur- 
rent Circuits (Continued) 13 

Resistivity and conductivity. Current density and voltage 
gradient. Kirchhoff's Laws. 

Chapter III. Conductors of Variable Cross-Section 22 

Current density and voltage gradient at a point. The radial flow 
of current. The resistance and conductance of irregular paths. 
The law of current refraction. 
Chapter IV. Representation of Alternating Currents and 

Voltages by Sine-waves and by Vectors 31 

Sinusoidal voltages and currents. Representation of a sine-wave 
by a vector. Addition and subtraction of vectors. Non-sinusoidal 
currents and voltages. 

Chapter V. Power in Alternating-current Circuits 45 

Power when current and voltage are in phase. The effective 
values of current and voltage. Some special methods for calculating 
the effective value of an irregular curve. Power when current and 
voltage are out of phase. 

Chapter VI. Inductance, Reactance, and Impedance 60 

Inductance as electromagnetic inertia. Reactance. Impedance. 
Influence of inductance with non-sinusoidal voltage. The extra 
or transient current in opening and closing a circuit. 

Chapter VII. Susceptance and Admittance 75 

Concept of susceptance. Concept of admittance. Equivalent 
series and parallel combinations. Impedances in parallel and admit- 
tances in series. 

Chapter VIII. The Use of Complex Quantities 82 

Addition and subtraction of projections of vectors. Rotation of 
vectors by ninety degrees. Impedance and admittance expressed 
as complex quantities or operators. 

Chapter IX. The Use of Complex Quantities (Continued) 91 

Power and phase displacement expressed by projections of vectors. 
Vectors and operators in polar coordinates. Vectors and operators 
expressed as exponential functions. 

ix 



x CONTENTS 

PAGE 

Chapter X. Polyphase Systems 99 

Two-phase system. Three-phase Y-connected system. Three- 
phase delta-connected system. 

Chapter XL Voltage Regulation of the Transformer 108 

Imperfections in a transformer replaced by equivalent resistances 
and reactances. The vector diagram of a transformer. Analytical 
determination of voltage regulation. — Approximate solution. Ana- 
lytical determination of voltage regulation. — Exact solution. 
Chapter XII. Performance Characteristics of the Induction 

Motor 122 

The equivalent electrical diagram of an induction motor. The 
analytical determination of performance. — Approximate solution. 
Starting torque, pull-out torque, and maximum output. 
Chapter XIII. Performance Characteristics of the Induction 

Motor (Continued) 133 

The secondary resistances and reactances reduced to the primary 
circuit. The circle diagram. The analytical determination of per- 
formance. — Exact solution. 

Chapter XIV. The Dielectric Circuit 143 

The electrostatic field. A hydraulic analogue to the dielectric 
circuit. The permittance and elastance of dielectric paths. Per- 
mittivity and elastivity of dielectrics. Dielectric flux density and 
electrostatic stress (voltage gradient). 

Chapter XV. The Dielectric Circuit (Continued) 157 

Energy in the electrostatic field. The permittance and elastance 
of irregular paths. The law of flux refraction. The dielectric 
strength of insulating materials. The electrostatic corona. Dielec- 
tric hysteresis and conductance. 
Chapter XVI. Elastance and Permittance of Single-phase 

Cables and Transmission Lines 171 

The elastance of a single-core cable. The elastance of a single- 
phase line. The influence of the ground upon the elastance of a 
single-phase line. The equations of the electrostatic lines of force 
and equipotential surfaces produced by a single-phase line. The 
elastance between two large parallel circular cylinders. 
Chapter XVII. Equivalent Elastance and Charging Current in 

Three-phase Lines 193 

Three-phase line with symmetrical spacing. Three-phase line 
with unsymmetrical spacing. 
Chapter XVIII. Dielectric Reactance and Susceptance in 

Alternating-current Circuits 203 

Dielectric reactance and susceptance. Current and voltage 
resonance. Voltage regulation of a transmission line, taking its dis- 
tributed permittance into account. Approximate formulae for the 
voltage regulation of a transmission line, considering its permittance 
concentrated at one or more points. 

Appendix 215 

Bibliography . . ... 219 



SUGGESTIONS TO TEACHERS 



(1) This book is intended to be used as a text in a course 
which comprises lectures, recitations, computing periods and 
home work. Purely descriptive matter has been omitted or only 
suggested, in order to allow the teacher more freedom in his 
lectures and to permit him to establish his own point of view. 
Some parts of the book are more suitable for recitations, others for 
reference in the designing room, others again as a basis for dis- 
cussion in the lecture rooni, or for brief theses. 

(2) Different parts of the book are made as much as possible 
independent of one another, so that the teacher can schedule 
them as it suits him best. Moreover, most chapters are written 
according to the concentric method, so that it is not necessary to 
finish one chapter before starting on the next. One can cover the 
subject in an abridged manner, omitting the last parts of some 
chapters. 

(3) The problems given at the end of nearly every article are 
an integral part of the book, and should under no circumstances 
be omitted. There is no royal way of obtaining a clear under- 
standing of the underlying physical principles, and of acquiring 
an assurance in their practical application, except by the solution 
of numerical examples. 

(4) The book contains comparatively few sketches, in order to 
give the student an opportunity to illustrate the important re- 
lations by sketches of his own. Making sketches, diagrams and 
drawings of electric circuits and machines to scale should be one 
of the important features of the course, even though it may not 
be popular with some analytically inclined students. Mechanical 
drawing develops precision of judgment, and gives the student a 
knowledge that is tangible and concrete. 

(5) The author has avoided giving definite numerical data, 
coefficients and standards, except in problems, where they are in- 
dispensable and where no general significance is ascribed to such 

xi 



Xll 



SUGGESTIONS TO TEACHERS 



data. His reasons are: (a) Numerical coefficients obscure the 
general exposition. (6) Sufficient numerical coefficients and de- 
sign data will be found in good electrical hand-books and pocket- 
books, one of which ought to be used in conjunction with this 
text, (c) The student is likely to ascribe too much authority to 
a numerical value given in a text-book, while in reality many 
coefficients vary within wide limits according to the conditions of 
a practical problem, and with the progress of the art. (d) Most 
numerical coefficients are obtained in practice by assuming that 
the phenomenon in question occurs according to a definite law, 
and by substituting the available experimental data into the cor- 
responding formula. This point of view is emphasized through- 
out the book, and gives the student the comforting feeling that he 
will be able to obtain the necessary numerical constants when 
confronted by a definite practical situation. 

(6) The treatment of the electrostatic circuit is made as 
much as possible analogous to that of the electrodynamic cir- 
cuit. The teacher will find it advisable to make his students 
perfectly familiar with the use of Ohm's law for ordinary electric 
circuits before starting on the electrostatic circuit. The student 
should solve several numerical examples involving voltages and 
voltage gradients, currents and current densities, resistances, re- 
sistivities, conductances and conductivities. He will then find 
very little difficulty in mastering the electrostatic circuit, and 
from these two the transition to the magnetic circuit, treated in 
the companion book, is very simple indeed. The following table 
shows the analogous quantities in the three kinds of circuits. 



Electrodynamic 
Voltage or e.m.f. 
Voltage gradient (or 
electric intensity) 

Electric current 
Current density 

Resistor 

Resistance 

Resistivity 

Conductor 

Conductance 

Conductivity 



Electrostatic 
Voltage or e.m.f. 
Voltage gradient (or 
electric intensity) 

Dielectric flux 
Dielectric flux density 

Elastor 

Elastance 

Elastivity 



Magnetic 
Magnetomotive force 
M.m.f. gradient (or mag- 
netic intensity) 

Magnetic flux 
Magnetic flux density 

Reluctor 

Reluctance 

Reluctivity 



Permittor (condenser) Permeator 

Permittance (capacity) Permeance 

Permittivity (dielec- Permeability 
trie constant) 



LIST OF PRINCIPAL SYMBOLS. 

The following list comprises most of the symbols used in the text. Those 

not occurring here are explained where they appear. When, also, a symbol 

has a use different from that stated below, the correct meaning is given where 

the symbol occurs. 

Page where defined 
Symbol. Meaning. or first used. 

a Radius of conductor of transmission line 176 

a Radius of core of cable 171 

A Cross-section 13 

b Interaxial distance between conductors of transmission line 176 

b Radius of inner surface of cable sheathing 171 

b Susceptance . 75 

C Constant 54, 72 

C Permittance or electrostatic capacity 147 

D Dielectric flux density 154 

D max Rupturing flux density 165 

e Electromotive force 1 

e Instantaneous value of voltage 34 

e Horizontal component of vector of e.m.f 83 

e' Vertical component of vector of e.m.f 83 

ei Local source of e.m.f 3 

et Terminal voltage 3 

E Effective value of alternating voltage 48 

E Vector of the voltage E 83 

E m Maximum value of voltage 34 

/ Frequency of alternating current or voltage 33 

F Mechanical force 60 

g Conductance 2 

g eq Equivalent conductance 8 

G Voltage gradient or electric intensity 16 

G max Rupturing voltage gradient 165 

h Elevation of conductor above the ground 181 

h Head of fluid 10 

h Instantaneous value of harmonic 54 

i Current „ . . . 1 

i Horizontal component of vector of current , 87 

i Instantaneous value of current 33 

i' Vertical component of vector of current , . 87 

/ Effective value of alternating current 48 

/ Vector of the current 1 88 

xiii 



XIV LIST OF PRINCIPAL SYMBOLS 

Page where defined 
Symbol. Meaning. or first used. 

Il Primary load current 116 

I m Maximum value of current ' . 34 

I m Mesh currents in squirrel-cage rotor 133 

j V - 1 .83, 85 

kb Breadth factor of winding „ . 133 

K Relative permittivity 151 

I Length 13 

log Common logarithm 172 

L Inductance 60 

Ln Natural logarithm 171 

m Mass . ... 60 

m Number of phases 133 

p Number of poles 135 

P Constant 73 

P Input per phase of induction motor 123 

P Power 10 

Pave Average power 48 

q Instantaneous displacement of electricity 193 

q Rate of discharge of a fluid 10 

Q Constant 73 

Q Quantity of electricity 144 

Q Quantity of heat 2 

r Resistance 1 

r ea Equivalent resistance 7 

R Resistance 8 

R Resistance at 0° C 5 

R t Resistance at t° C 5 

s Slip of induction motor 123 

S Area of curve 53 

S Elastance 148 

t Time 33 

T Temperature 6 

T Time of one cycle of alternating wave , 33 

u Variable angle 33 

U Current density 15 

v Velocity 60 

V Volume 158 

W Energy 46 

W Density of energy 158 

x Reactance 63 

x Variable radius 171 

y Admittance 76 

y Ordinate of curve 50 

Y Admittance operator 89 

z Impedance 67 

Z Impedance operator 88 



LIST OF PRINCIPAL SYMBOLS XV 

Page where defined 

Symbol. Meaning. or first used. 

a Angle 43, 94 

a A ratio 189 

a Temperature coefficient . . . .' 5 

7 Conductivity 14 

e Base of natural system of logarithms 72 

Difference of temperature 2 

Phase angle 82 

0i Angle of incidence of current 28 

0i Angle of incidence of dielectric flux 163 

02 Angle of refraction of current 28 

2 Angle of refraction of dielectric flux 163 

k Permittivity 151 

K a Permittivity of air 151 

p Resistivity 13 

a Circle coefficient or dispersion factor of induction motor 138 

<r Elastivity 152 

a a Elastivity of air 152 

r Time constant of a circuit 72 

<f> Phase angle 34 

4> Magnetic flux 114 

m Maximum value of magnetic flux 114 

\f/ Angle 121 

co Angle 52 

O Angle 52 



THE ELECTRIC CIRCUIT 



CHAPTER I 



FUNDAMENTAL ELECTRICAL RELATIONS IN DIRECT- 
CURRENT CIRCUITS 

1. The Volt, the Ampere, the Ohm, and the Mho. The 

student is supposed to be familiar with Ohm's law, both theoret- 
ically and from his laboratory experience. A brief synopsis of 
the law, given below, is intended to refresh the relations in his 
mind, and to establish a point of view which permits of extending 
these relations to alternating-current circuits. Moreover, the 
law is presented in a form applicable to magnetic and dielectric 
circuits. 

When the current in a conductor is steady and there are no 
local electromotive forces within the conductor, the value of the 
current is proportional to the voltage between the terminals of 
the conductor. This is an experimental fact, called Ohm's law. 
The word " conductor" is used here in the sense of "the part of 
the circuit under consideration. " It may consist of two or more 
distinct physical conductors. Considering the electromotive force 
e as the cause of the current i, this law merely states that the effect 
is proportional to the cause, or 

e = r>i, , . . (1) 

where the coefficient of proportionality r is called the resistance 
of the conductor. When the current is expressed in amperes, 
and the electromotive force in volts, the resistance r is measured 
in units called ohms. 

Ohm's law is sometimes written in the form 

i = 9-e, (2) 

1 



2 THE ELECTRIC CIRCUIT [Art. 1 

where the coefficient of proportionality 

<7 = l/r (3) 

is called the conductance of the conductor. The reason for this 
name is easy to see: The resistance r shows how difficult it is to 
force a unit current through a given conductor, while its recip- 
rocal g shows how easy it is to produce the same current in the 
same conductor. Conductances are measured in units called 
mhos, one mho being the reciprocal of one ohm. Hence, a resist- 
ance of one ohm represents at the same time a conductance of 
one mho; a resistance of two ohms has a conductance of one-half 
mho, etc. Increasing the resistance of a winding from 4 to 5 ohms 
reduces its conductance from 0.25 to 0.20 mho. 

It will be seen below that in some problems it is convenient to 
use conductances instead of resistances. Both are fundamental, 
and there is no reason why Ohm's law should not have been ex- 
pressed originally by eq. (2) instead of (1). 

With our present meager knowledge of the true nature of 
electrical phenomena, it is well-nigh impossible to give a clear 
physical meaning of the quantities under discussion without 
resorting to analogies. For instance, the flow of current through 
a conductor may be compared to the flow of heat through a 
rod; the voltage or difference of electric potential is analogous 
to the difference of temperature 6 at the ends of the rod, and the 
electric current to the quantity of heat Q passing through a cross- 
section of the rod in unit time (the rate of flow of heat). The 
ratio of to Q is sometimes called the thermal resistance of 
the rod. 1 

Again, the phenomenon of the flow of electricity is somewhat 
analogous to the flow of water through pipes. The hydraulic 
head may be likened to the voltage, and the rate of discharge of 
water to the current. With very low velocities, in capillary 
tubes, the discharge is proportional to the head, so that eqs. (1) 
and (2) hold true for the flow of water. 

Whatever the reasons which have led originally to the choice 
of the magnitudes of the ampere, the ohm, and the volt, these 
units may be considered at present, for all practical and most 
theoretical purposes, as arbitrary units, like the foot, the pound, or 

1 It even has been proposed to measure this resistance in thermal ohms or 
thohms. 



Chap. I] DIRECT-CURRENT CIRCUITS 3 

the meter. Their values have been established by an interna- 
tional agreement, whence the name, international electrical units. 
These units are represented by concrete standards with minutely 
specified dimensions and properties; the ohm by a column of 
mercury, the ampere by a silver voltameter, and the volt by a 
standard cell. It is understood, of course, that only two out of 
the three units need to be standardized, the third being determined 
either as their product, or their ratio. It has been decided by 
international agreement to consider the ampere and the ohm as 
fundamental units, the volt being derived from them. Hence 
the present system of practical electrical units is properly called 
the ampere-ohm system. This fact does not preclude, of course, 
the use of standard cells as secondary standards. 

The ampere, the volt, and the ohm are connected by simple 
multipliers (powers of 10) with the absolute electromagnetic units 
(the C.G.S. system of units). It is conceded at present by some 
prominent physicists that the choice of the units was not quite 
fortunate, according to our present understanding of the electro- 
magnetic relations. Since, however, it is too late to change these 
units, it is better to consider them as arbitrary, and not con- 
nected in any way with the magnitudes of the centimeter, the 
gram, and the second. 

In applying Ohm's law to practical problems, it must be 
clearly remembered that e represents the net voltage acting be- 
tween the ends of the conductor r. This is important when the 
circuit contains sources of counter-electromotive force, such as 
electric batteries, or motors. Let, for instance, the total resistance 
of a circuit, connected across the terminals of a generator, be 12 
ohms, and let the terminal voltage of the generator be 120 volts. 
Then the current is equal to 10 amperes, provided that there are 
no counter-electromotive forces in the circuit. Let, however, the 
circuit contain a storage battery of, say, 24 volts, connected so 
as to be charging, that is, opposing the applied voltage. The 
current in the circuit is now only (120 — 24)/12 = 8 amp., the 
value 120 — 24 = 96 being the net voltage in the external circuit. 
Should the terminals of the battery be reversed, so as to help the 
generator voltage, the current would increase to (120 + 24)/ 12 
= 12 amp. 

Thus, when there is an external or local source of electro- 
motive force, say e h within a conductor, the terminal voltage e t 



4 THE ELECTRIC CIRCUIT [Art. 1 

between the ends of the conductor is added algebraically to ei, 
so that we have, instead of eq. (1), 

e t + ei= i»r, ,. (4) 

where ei is considered positive when in the same direction as e t . 
In the foregoing numerical example the counter-e.m.f . is therefore 
considered negative. 

In numerical computations it is sometimes convenient to use 
multiples and submultiples of the units originally agreed upon, 
in order to avoid large numbers or very small fractions. This is 
accomplished by adding to the names of the original units certain 
Greek prefixes for the multiples, and Latin prefixes for the sub- 
multiples. These prefixes are as follows: 

deca .... ten deci one tenth 

hecto . . . one hundred centi one hundredth 

kilo one thousand milli one thousandth 

mega. . .one million micro one millionth. 

For instance, instead of 10,000 amperes one may say or write 
10 kiloamperes; instead of 0.0003 volt one may say 0.3 millivolt, 
or 300 microvolts, etc. Another way to avoid very large or very 
small numbers is to use 10 to the proper power as a multiplier. 
For instance, one may speak of a resistance of 7 X 10~ 6 ohm, of a 
conductance equal to 5 X 10 7 mhos, etc. 

Prob. 1. In order to determine the resistance of the armature of an 
electrical machine, a direct current is sent through it and the drop of volt- 
age is measured between the brushes. The following are the readings: 



Volts 

Amperes 


0.44 

8.1 


0.73 
12.9 


1.00 
18.1 


1.33 
24.0 


1.73 
31.0 



What is the most probable value of the resistance? Hint: Take an 
average of the ratios, or better, plot the volts against the amperes as 
abscissae and draw a straight line through the origin. 

Ans. 0.0559 ohm. 

Prob. 2. The resistance of a transmission line is 1.2 ohms. What 
voltage is necessary at the generating end in order to produce a current 
of 75 amp. (a) when the line is short-circuited at the receiving end; 
(b) when a pressure of 500 volts must be maintained at the receiving 
end? Ans. 90 volts; 590 volts. 

Prob. 3. The armature resistance of a 250-volt generator is 0.025 
ohm. At what current will the voltage drop in the armature be equal 
to 4 per cent of the terminal voltage? Ans. 400 amp. 



Chap. I] 



DIRECT-CURRENT CIRCUITS 



Prob. 4. The conductance of a bath of molten metal is 5 kilomhos; 
what voltage is required to send a current of 7 X 10 4 amp. through it? 

Ans. 14 volts. 

Prob. 5. The coil of a regulating electromagnet of 50 ohms resistance 
is connected across a 110- volt line; the voltage of the line fluctuates by 
± 10 per cent. In order to make the regulating mechanism more sensi- 
tive, that is, in order to accentuate the fluctuations of current in it, a 
counter-e.m.f. storage battery of negligible resistance is connected in 
series with the coil. What must be the voltage of the battery if the cur- 
rent in the coil at 120 volts must be twice that at 100 volts? 

Ans. 80 volts. 

2. Temperature Coefficient. The resistance of all metals and 
of practically all alloys increases with the temperature, according 
to a rather complicated law. Within the usual limits of tem- 
perature the increase in resistance is nearly proportional to the 
temperature rise; in other words, the relation between the resist- 
ance of a conductor and its temperature is represented by a straight 




Fig. 1. The relation between the resistance and the temperature of 

metals. 

line MN (Fig. 1). Let the resistance at 0° C. be R ohms; then 
the resistance at some temperature f C. is 

R t = R Q (1 -1- at), (5) 

where a is called the temperature coefficient of the material. For 
the values of a for various materials see an electrical handbook. 
For the most important material, copper, a = 0.0042; in other 
words, the resistance of a copper conductor increases by 0.42 per 
cent for each degree centigrade, considering the resistance at 
0° C. as 100 per cent. 

The formula given below is sometimes more convenient in 
computations than formula (5). Assume that the same straight- 



6 THE ELECTRIC CIRCUIT [Art. 2 

line law (Fig. 1) holds for low temperatures, and let temperatures 
be measured from the point A at which the straight line crosses 
the axis of abscissae. Denoting temperatures from this point by 
T, we have, for any two temperatures, 

R1/B2 = T,/T 2 (6) 

With this formula it is not necessary to refer computations to 
the resistance at 0° C. The point A is found from the condition 
R t = 0, from which, according to eq. (5), Ia = — 1/a. Thus, for 
any temperature, 

T = t+.l/a. (7) 

For copper, T = t + 238.1; that is, point A lies 238.1° C. below 
the freezing point of water. This does not mean that the resist- 
ance of copper actually varies according to this law at such 
temperatures; A is merely a fictitious point through the position 
of which it is convenient to express the equation of the full-drawn 
part of the straight line in Fig. 1. 

Let, for instance, the resistance of the winding of an electric 
machine be 0.437 ohm at the room temperature of 22° C. After 
the machine has been run for several hours the resistance of the 
same winding is found to be 0.482 ohm, with the room temper- 
ature unchanged. Let it be required to calculate the final tem- 
perature of the winding from the increase in its resistance. We 
have 7\ = 238.1 + 22 = 260.1, and according to eq. (6) the un- 
known final temperature T 2 = 260.1 X (482/437) = 286.9, or 
h = 286.9 - 238.1 = 48.8° C. Two other practical formulae for 
temperature rise will be found in Appendix E of the Standard- 
ization Rules of the American Institute of Electrical Engineers. 
These rules are reprinted in most electrical handbooks and pocket- 
books. See also a convenient method given in problem 2 below. 

Prob. 1. The resistance of a conductor increases by 31 per cent from 
23° to 75° C. What is a in formula (5)? Ans. 0.00691. 

Prob. 2. The relation between the resistance and the temperature 
of copper conductors is easily obtained on an ordinary slide rule, as fol- 
lows: On the lower movable scale mark 0° C. on division 238; 10° C. on 
division 248, and so on. Set a known resistance on the lower fixed scale, 
and bring the corresponding temperature opposite. Then the resistance 
at any other temperature is read opposite the corresponding division on 
the temperature scale. Give an explanation of this method. 

Prob. 3. Prove the formulae for Ri+r and r in the above-mentioned 
Standardization Rules. 



Chap. I] DIRECT-CURRENT CIRCUITS 7 

3. Resistances and Conductances in Series and in Parallel. 
When resistances are connected in series, the total resistance of 
the circuit is increased. This can be more easily seen by resort- 
ing to analogies. For instance, if the length of a pipe carrying a 
fluid be increased, the frictional resistance to the flow becomes 
greater; in like manner, a long rod offers a more difficult path for 
the passage of heat than a short one. In the electric circuit, the 
equivalent resistance of two conductors in series is equal to the 
sum of their individual resistances, as is shown below. This follows 
from the experimental fact that electricity in its flow behaves like 
an incompressible fluid; that is, the same quantity of it must pass 
in a given time interval through all the cross-sections of a circuit. 

Let two conductors, r± and r 2 , be connected in series across a 
source of voltage e, and let a current i flow through them. Part 
of the total voltage e is spent in overcoming the resistance of the 
first conductor, the rest in overcoming that of the second con- 
ductor. But, according to Ohm's law, when the conditions are 
steady, the voltage across the first conductor, e\ = i«rij the 
voltage across the second is e 2 — i • r 2 . Adding these two equa- 
tions gives the total voltage 

e = ei + e 2 = i{n + r 2 ). 

An equivalent resistance, r eq , by definition, is one which, with 
the same total voltage e, allows the same current i to pass through 
the circuit, as the combination of the given conductors. Hence, 

Comparing the two foregoing equations gives 

r eq = n + r 2 (8) 

The law is true for any number of conductors in series; it may 
be proved by successively combining them into groups of two. 

When several conductors are connected in parallel, the voltage 
across them is common to all the branches, so that we have 



e = i\*ri 
e = i 2 • r 2 



(9) 



where i h i 2j . . . are the currents in the separate branches. The 
total current is equal to the sum of the currents in the separate 



8 THE ELECTRIC CIRCUIT [Abt. 3 

branches, because electricity behaves in its flow like an incom- 
pressible fluid. Thus, the equivalent resistance, r eq , is determined 
by the condition 

e = (ii + ii + etc.) • r eq . (10) 

Substituting the values of ii, i 2 , etc., from (9) into (10) and 
canceling e, gives 

l/r eq = 1/ri + l/r 2 + etc., (11) 

or, in words: when two or more conductors are connected in 
parallel, the reciprocal of the equivalent resistance is equal to the 
sum of the reciprocals of the individual resistances. 

We have defined conductance as the reciprocal of resistance, 
so that eq. (11) may be written also in the form 

g eq = 01 + #2 + etc (12) 

It will thus be seen that it is convenient to use conductances in 
parallel circuits and resistances in series circuits. The simple 
rule is: Resistances are added in series; conductances are added in 
parallel. This rule follows directly from the physical concept of 
resistance and conductance. 

Prob. 1. Prove that when two conductors, 1 and 2, are in parallel 

t'l/ii = gi/V2 = r*/n, ."""■.- . . . . . (13) 
and that when they are in series 

ei/e-i = ri/r 2 = QilQx (14) 

Prob* 2. Show that when two resistances are in parallel the equivalent 
resistance 

r eq = rir 2 /(ri + r 2 ), (15) 

and that for two conductances in series 

g eq = giQ2/{gi + Ql) (16) 

Prob. 3. Two resistances, n = 5 ohms and r 2 = 7 ohms, are connected 
in series. Resistance r x is shunted by a comparatively high resistance 
Ei = 100 ohms; r 2 is shunted by a resistance R% = 50 ohms. What is 
the equivalent resistance of the whole combination? Solution: 

Equivalent conductance of n and Ri is 0.2 + 0.01 = 0.21 mho; 
Equivalent resistance of n and R i is 1/0.21 = 4.76 ohms; 

Equivalent conductance of r 2 and R 2 is 

0.1429 + 0.0200 = 0.1629 mho; 
Equivalent resistance of r 2 and R 2 is 1/0.1629 = 6.14 ohms. 

Ans. 4.76 + 6.14 = 10.90 ohms. 

Prob. 4. Four resistances, n = 1.2, r 2 = 1.7, R = 25, and r = 
750 ohms, are connected as shown in Fig. 2. The generator voltage 



Chap. I] 



DIRECT-CURRENT CIRCUITS 



9 



between the points A and B is 500 volts. Determine the current through 
the resistance R and the voltage across this resistance. 1 Solution: Com- 
bine the resistances r<i and R into one; determine the conductance 
l/(R + r 2 ), and combine it with the leakage conductance l/r . De- 



Ao 



^VWWWVNA 



Bo- 




N D 

Fig. 2. A series-parallel combination of resistances. 



termine the equivalent resistance between the points M and N. and the 
total resistance between A and B. Having found the total current, sub- 
tract from the generator voltage the voltage drop in the part AM of the 
line. This will give the voltage across MN, and consequently the value 
of the leakage current. After this, 
the drop in r 2 is determined, and 
thus the voltage across the resist- 
ance R is found. 
Ans. 447.3 volts; 17.8 amps. 
Prob. 5. The armature winding 
of a direct-current machine (Fig. 3) 
consists of 108 coils; the conduct- 
ance of each coil is 61 mhos. The 
coils are connected in series in such 
a way that the circuit is closed 
upon itself. Two positive and two 
negative brushes are placed alter- 
nately at four equidistant points of 
the winding, so as to divide it into 
four branches in parallel. The two 
positive brushes are connected to- 
gether, as are also the two negative 
brushes. What is the equivalent 
resistance of the armature between 
the terminals of the machine? 

Ans. 0.1106 ohm. 2 




Terminals 



Fig. 3. A four-pole multiple winding. 



1 This combination represents a transmission line the resistance of which 
is rjL + r 2 ; the useful load resistance is represented by R, and the leakage re- 
sistance by r . The problem in a generalized form is of great importance in 
the theory of alternating-current circuits (see Figs. 41 and 42). 

2 For details of armature windings see the author's Experimental Electrical 
Engineering, Vol. 2, chap. 30. 



10 THE ELECTRIC CIRCUIT [Art. 4 

Prob. 6. Two equal resistances of r ohms each are connected in 
series. When one of them is shunted by an unknown resistance R, the 
total resistance of the combination decreases by 10 per cent. Find the 
value of R. Ans. 4 r. 

4. Electric Power. The electric power (energy per unit 
time) converted into heat in a conductor is found by experi- 
ment to be proportional to the resistance of the conductor and 
to the square of the current (Joule's law). The practical unit of 
power, the watt, is so selected that the coefficient of proportion- 
ality is unity, or the power 

P = i 2 r = i 2 /g. ....... (17) 

Either i or r may be eliminated from this expression, using Ohm's 
law. This gives three more expressions for power: 

p = e • i = e 2 /r = e 2 g (18) 

All these expressions are used in practice, depending upon which 
quantities are known in a particular case. 

The expression e • i is the fundamental one; it is analogous to 
the expression h • q for the power lost by friction in a pipe in which 
a fluid is in uniform motion. In the pipe, the energy lost per 
unit time is equal to the rate of discharge q times the head h lost 
in friction; in other words, it is equal to the quantity factor 
times the intensity factor. In an electric circuit the current i 
is the quantity factor, while the voltage drop e is the intensity 
factor. 

If P in eqs. (17) and (18) is expressed in kilowatts, or in mega- 
watts, a numerical factor equal to 10 -3 or 10~ 6 respectively is 
introduced on the right-hand side of the equation. Sometimes 
the output of a motor is measured in horse-power; the English 
horse-power is equal to 746 watts, while the metric horse-power 
is 736 watts. It is strongly recommended by the International 
Electrotechnical Commission that the odd and superfluous unit 
"horse-power" be dropped altogether and that mechanical power 
be expressed also in watts (or kilowatts). This means that elec- 
tric motors as well as generators should be rated in kilowatts. 

Sometimes the duty of a machine is expressed in kilogram- 
meters per second; the conversion ratio to watts is: 

1 kg.-m. per second = 9.806 watts. 



Chap. I] DIRECT-CURRENT CIRCUITS 11 

In many cases, however, it is not necessary to introduce either 
kilogram-meters or calories, since mechanical, thermal, and electri- 
cal energy can all be expressed in joules (watt-seconds). 

If a conductor contains a local e.m.f. ei, the power commu- 
nicated to this part of the circuit, between its terminals, is 
equal to e t • i, where e t is the terminal voltage. But the power 
i 2 r converted into heat may be either smaller or larger than e t i y 
depending upon the polarity or direction of e L . Multiplying both 
sides of eq. (4) by i, we find that the power 

p = et i + e t i = i 2 r (19) 

Let e t be positive, that is, in the same direction as e t (for instance, 
an extra battery or generator connected into the circuit to boost 
the voltage); the power i 2 r converted into heat is in this case 
larger than e t i, because the power supplied by the local source 
of e.m.f. is also converted into heat. If, however, ei is negative, 
that is, if it acts as a counter-e.m.f. (which is usually the case in 
practice), the power converted into heat is smaller than e t i. In 
this case the power eti is communicated to the local source of 
e.m.f. If this source is a storage battery, the energy is stored in 
chemical form, and may be made available at a later time; if it 
is a motor, the energy is converted into mechanical work on the 
motor shaft. Let, for instance, the voltage at the terminals of a 
circuit be 110 volts, and let the counter-e.m.f. of a motor in the 
circuit be 100 volts; assume the current through the circuit to 
be 20 amp. Then the voltage drop due to resistance in the 
conductors is only 10 volts, and the power converted into heat 
is 200 watts. The power communicated to the motor is 2000 
watts, and the total power supplied to the circuit is 2200 watts. 

The unit of electrical energy is the watt-second, or joule. When 
the heat dissipated in a conductor must be expressed in thermal 
units, use the relation 

1 kg.-calorie = 4186 joules. 

Prob. 1. The armature current of a 220-volt direct-current motor at 
a certain load is 63 amp., and the armature resistance is 0.14 ohm. How 
much electric power is converted into mechanical form, and what is the 
torque developed by the armature if the speed is 1050 r.p.m.? 

Ans. 13.3 kw.; 12.3 m.-kg. 

Prob. 2. If the currents in the shunted resistances Ri and R% (problem 
3, Art. 3) represent pure loss of power, what is the efficiency of the whole 
arrangement? Solution: Let the voltage across the resistances ri and 



12 THE ELECTRIC CIRCUIT [Art. 4 

Ri be e. Then the voltage across r 2 and R 2 is e • (6.14/4.76) = 1.29 e. 
Hence, the useful power is e 2 /5 + (1.29 e) 2 /7 = 0.438 e 2 watts. The power 
lost in the resistances 22 1 and R 2 is e 2 /100 + (1.29 e) 2 /50 = 0.0433 e 2 watts. 
The efficiency is 43.8/(43.8 + 4.33) = 90 per cent. 

Prob. 3. The heating element of a 110-volt electric kettle must be 
designed so that it will heat 1.5 liters (1 liter = 1 cu. decimeter) of water 
at a rate of 10° C. per minute, assuming no losses by radiation. What 
are the resistance of the element and the rated current of the utensil? 

Ans. 11.6 ohms; 9.5 amp. 

Prob. 4. It is required to calculate the exciting current i, the number 
of turns n per pole, and the resistance r per turn of a field coil of a 5000-kw. 
6-pole turbo-alternator, from the following data : The excitation required 
at the rated load is 9000 amp .-turns per pole; at short overloads 12,000 
amp.-turns per pole are needed. The external area of the field coil is 
280 sq. dm. ; in continuous service, 4 sq. cm. of cooling surface must be 
allowed per watt converted into heat, in order to avoid overheating the 
coils. The exciter voltage is 125, and during the overload about 10 per 
cent of this voltage must be absorbed in the field rheostat, as a margin. 
Hint: Solve the following three equations; in = 9000; i 2 rn = 28,000/4; 
(l2,000/n)nr = 0.9 X 125/6. 

Ans. 500 amp.: 18 turns; 1.562 X 10~ 3 ohms. 



CHAPTER II 

FUNDAMENTAL ELECTRICAL RELATIONS IN DIRECT- 
CURRENT CIRCUITS— (Continued) 

5. Resistivity and Conductivity. A cylindrical conductor 
may be considered as a combination of unit conductors in series 
and in parallel. For instance, a wire 12 m. long and having a 
cross-section of 70 sq. mm. may be regarded as composed of 70 X 
12 = 840 unit conductors, each of one square millimeter cross- 
section, and one meter long. These unit conductors are first 
combined into sets of 70 in parallel, and then the 12 sets are 
connected in series. The resistance of such a unit conductor, 
made of copper, and at a temperature of 0° C, is about 0.016 
ohm. A set of 70 unit conductors in parallel has T V of the resist- 
ance of one, because the current is offered 70 paths, instead of 
one; twelve sets connected in series offer twelve times the resist- 
ance of one set. Therefore, the resistance of the given conduc- 
tor is (0.016/70) X 12 = 0.002743 ohm. 

Each material is characterized by the resistance of a unit con- 
ductor made out of it. The resistance of a unit conductor at a 
specified temperature is called the resistivity 1 of the material and 
is denoted by p. Thus, the resistance of a conductor of a length I 
and cross-section A is 

r = p-l/A. ....... (20) 

The numerical value of p depends upon the units of length and 
resistance used. A unit conductor may, for instance, have a 
cross-section of one square millimeter, and may be one meter, or 
one kilometer long; or it may be a centimeter cube. In the 
English system it may have a cross-section equal to one circular 
mil, and a length of one foot, one thousand feet, one mile, or any 
such specified length. Besides, the resistivity may be expressed in 
ohms, megohms, or microhms. In each case, the unit of resistance 
and the units in which the dimensions of I and A are expressed in 
1 The older name is " specific resistance." 
13 



14 THE ELECTRIC CIRCUIT [Art. 5 

formula (20) are selected so as to suit the convenience of the 
user of the formula. For the values of p for different materials 
see one of the various handbooks and pocketbooks for electrical 
engineers. 

In some cases it is more convenient to use the conductance of 
the unit conductor, instead of its resistance. The conductance 
of a unit conductor, at the specified temperature, is called the 
electric conductivity (or specific conductance) of the material; the 
conductivity is the reciprocal of the resistivity of the samejnate- 
rial. Denoting this conductivity by y, we have y = 1/p. By 
reasoning similar to that given above, we find that the conduc- 
tance of a conductor having the dimensions I and A is 

g = v-A/l (21) 

Prbb. 1. The resistivity of aluminum equals 2.66 microhms per cubic 
centimeter; what is its conductivity per mil-foot? Solution: The re- 
sistance of a conductor one foot long and having a cross-section equal to 
one circular mil is 2.66 X 10" 6 X 197,300 X 30.48 = 16 ohms; where 
197,300 = (1000/2.54) 2 X 4/tt is the factor for converting square centi- 
meters into circular mils, and 30.48 is the number of centimeters in one 
foot. Ans. 0.0625 mho per circular mil-foot. 

Prob. 2. Each field coil of an electric machine has 720 turns, the 
average length of a turn being about 1.5 m. What size wire is required 
if the hot resistance of the coil is to be 1.14 ohms? According to the 
A. I. E. E. Standardization Rules, a temperature rise of 50° C. is allowed 
above the air at 25° C. Ans. About 20 sq. mm. 

Prob. 3. A given current i is to be transmitted at a given voltage 
between two given localities whose distance apart is I. Deduce an 
expression for the most economical size of the line conductor. A small 
conductor means a saving in the original investment, but a higher operat- 
ing cost on account of the power lost in the conductor, and vice versa. 
The most economical conductor is one for which the annual interest and 
depreciation plus the cost of the i 2 r loss per year is a minimum. Solu- 
tion : Let the cost of one watt-year be p cents, and let the conductor cost 
q cents per cubic centimeter, installed. Let 5 be the annual interest 
and depreciation in per cent to be allowed on the original cost of the con- 
ductor. The cost of the power lost in the line is pi 2 pi/ A, and the initial 
cost of the conductor is qlA. The condition of the problem is that 

pi 2 P l/A + dqlA + K = min., (22) 

where the constant K represents the interest and depreciation on the 
poles, insulators, etc., the size of which is essentially independent of the 
size of the conductor. Equating the first derivative with respect to A to 
zero, we get — pi 2 p/A 2 -\- 8q = 0, or 

pi 2 P /A = 8qA (22a) 



Chap. II] DIRECT-CURRENT CIRCUITS 15 

In other words, the most economical cross-section is that for which the 
sum charged to the annual interest and depreciation is equal to the cost 
of the wasted energy. This result is independent of the length of the 
line and of the voltage, and is known as Kelvin's law of economy. 1 Know- 
ing all the data, the cross-section A can be calculated from condition 
(22a) ; see also problems G and 7 in Art. 6. 

Prob. 4. A transmission line from the generating station A to a place 
B is I kilometers long. At B the line is divided into two branches; one 
to C, 1 1 km. long, and carrying a current i\; the other to D, l 2 km. long, 
and carrying a current i 2 . The total permissible voltage drop from A 
to either C or D is e volts (one way). Determine the sizes of the con- 
ductors in the three parts of the line so as to make the total initial cost 
of copper a minimum. Solution: Take the unknown voltage drop x 
from A to B as the independent variable; then the three cross-sections 
are determined by the conditions ipl/A = x, iiph/Ai = e — x, and 
iipU/A-i = e — x. The value of x itself is determined by the condition 
that I A + hAx + UAi = min. Substituting the values of A, Ai and A 2 
into this expression, and equating the first derivative with respect to x 
to zero, we get il 2 /x 2 = iih 2 /(e - x) 2 + W/i* - x) 2 . Extracting the 
square root of each member of this equation and solving for x, we find that 



1 + [(ii/i) (Zi/0 2 + (t*/i) (h/D 2 V 



Having found x, the three cross-sections are easily calculated from the 
three conditions written above. 

6. Current Density and Voltage Gradient. When a current 
is distributed uniformly over the cross-section of a cylindrical 
conductor, it is convenient to speak of the current density, or the 
current per unit cross-section of the conductor. Denoting this 
density by U, we have 

U = i/A (23) 

U is measured in amperes (or kiloamperes, milliamperes, etc.) 
per square centimeter, or per square millimeter. The current 
density is numerically equal to the current through each unit 
conductor of which the given conductor consists. 

1 In practice the selection of the cross-section of a line conductor is 
determined by many other considerations besides that of economy, as out- 
lined above. For instance, it may be desired to reduce the original invest- 
ment to a minimum while the load is small, and to change the conductors 
to a larger size afterwards. The problem above is intended only to introduce 
the reader into this subject. He will find numerous contributions treating 
of more complicated cases in various periodicals and transactions. See also 
A. C. Perrine, Electrical Conductors, Chapter 8. 



16 THE ELECTRIC CIRCUIT [Art. 6 

When the voltage drop is distributed uniformly along a con- 
ductor, it is convenient to speak of voltage drop per unit length. 
This voltage drop across a unit length is called the voltage gradient, 
and is measured in volts, kilovolts, millivolts, etc., per meter or 
per centimeter. Denoting the voltage gradient by G, we have 

G = e/l (24) 

The value of G characterizes the electrical condition at a point 
(or a cross-section) of the conductor; for this reason G is some- 
times called the electric intensity at a point. 

Having previously introduced the resistance p and the conduct- 
ance 7 of a unit conductor, we can now write Ohm's law for the 
unit conductor, in the form 

G= P U = U/y (25) 

Equation (25) has a definite meaning also without the concept 
of the unit conductor; namely, it gives the relation between the 
voltage gradient and the current density at a point, for a given 
material. The reader can easily think of a thermal analogue. 
Hooke's law for elastic materials is also somewhat analogous to 
eq. (25), because it expresses a straight-line relation between the 
cause and the effect. Equation (25) can be deduced directly 
from eq. (1) by writing the latter in the form Gl = (pi/ A) • UA 
and canceling I and A. 

Prob. 1. What is the voltage drop per kilometer of a copper wire 
having a cross-section of 70 sq. mm., and carrying a current of 150 amp.? 
Solution: U= 150/70= 2.143 amp. per sq. mm. The conductivity of 
copper 7, at the temperature of the line, is equal to 57 mhos for a unit 
conductor of one square millimeter cross-section and one meter long. 
Therefore, the electric intensity or the voltage drop per meter of length, 
according to formula (25), is 2.143/57= 0.0376 volt per meter. 

Ans. 37.6 volts/km. 

Prob. 2. What is the expression for power converted into heat in a 
unit conductor? Ans. 

p = G-U =U 2 P = U 2 /y = y • G 2 (25a) 

Prob. 3. What is the amount of power lost in the conductor con- 
sidered in problem 1? 

Ans. (0.0376 X 2.143) X 70 X 1000 = 5640 watts. 

Prob. 4. The space available on the frame of a generator for a rec- 
tangular field coil is 16 X 12 cm. for the inside dimensions, and 28 X 
24 cm. for the outside dimensions; the limiting height is 15 cm. What 
current density can be allowed in the coil, if 12 sq. cm. of exposed surface 
are required per watt loss, in order that the temperature of the coil shall 



Chap. II] DIRECT-CURRENT CIRCUITS 17 

not exceed the safe limit? The space factor of the coil is 0.55; in other 
words, 55 per cent of the gross space is occupied by copper, the rest being 
taken by the air spaces and the insulation. Solution: The exposed 
surface is 2(28 + 24) X 15 = 1560 sq. cm.; therefore, 130 watts loss 
can be allowed in the coil. With a space factor of 0.55 the useful cross- 
section of copper is 6 X 15 X 0.55 = 49.5 sq. cm. ; the average length 
of one turn equals 2(22 + 18) = 80 cm. Therefore, the coil contains 
4950 X 0.8 = 39G0 unit conductors, each one meter long and one square 
millimeter in cross-section. The permissible loss per unit conductor is 
130/39G0 = 0.0328 watt. Hence, according to the answer to problem 2 
above, U = Vo.0328 X 57 = 1.37 amp. per sq. mm. This result is 
independent of the size of the wire, as long as the space factor remains 
approximately the same. The maximum ampere-turns are 137 X 49.5 = 
6780, and, for a constant space-factor, are also independent of the size 
of the wire. 

Prob. 5. What are the size of wire and the exciting current in the pre- 
ceding problem if the voltage drop must not exceed 20 volts at 80° C? 

Ans. 57 sq. mm. ; 78 amp. 

Prob. 6. Referring to problem 3 in Art. 5, what is the gene ral expr es- 
sion for the most economical current density? Ans. U = v8q/(pp). 

Prob. 7. Referring to the preceding problem, what is the most eco- 
nomical current density if copper costs 15 cents per pound, the annual 
interest and depreciation is taken at 12 per cent, and the estimated cost 
of wasted power is 22 dollars per kilowatt-year? 

Ans. 0.95 amp. per sq. mm., taking p at 25° C. 

7. Kirchhoff's Laws. Consider an arbitrary network of 
conductors (Fig. 4), with sources of e.m.f. connected in one or 
more places. When such a system is left to itself, definite cur- 
rents will flow through the conductors, and definite differences of 
potential will be established between the junction points of the 
conductors. Thus, if all the resistances and e.m.fs. are given, it 
ought to be possible to calculate the magnitude and direction of 
all the currents. The distribution of the currents is such that 
two conditions are satisfied: 

(1) As much current flows toward each junction as from it, 
because electricity behaves like an incompressible fluid. For any 
junction this is expressed mathematically by the equation 

2i = 0, (26) 

in which all the currents flowing toward the junction are taken 
with the sign plus, all those flowing away from it with the sign 
minus, or vice versa. Thus, for instance, at the point C let the 
currents flowing toward the junction be 20 and 30 amp. respec- 



18 



THE ELECTRIC CIRCUIT 



[Art. 7 



tively, and one of the currents flowing away from it be 40 amp. 
Then the fourth current must necessarily be 10 amp. flowing 
away from C, because 20 -+ 30 - 40 - 10 = 0. Equation (26) 
is called Kirchhoff's first law. 

(2) The sum of the terminal voltages along any closed circuit 
in the network is equal to zero, or 

2e« = 0. . . , (27) 

Consider, for instance, the path ABCDEFA, and connect a zero- 
center voltmeter first between A and B, then between B and C, 
and so on, every time transferring both terminals, so that one 




Fig. 4. A network of conductors, illustrating Kirchhoff's laws. 



particular terminal of the instrument always leads the other. 
Consider the deflections to one side of the zero point as positive, 
to the other side as negative. Equation (27) means that the 
algebraic sum of these readings is equal to zero. The reason is 
as follows: The reading A -B shows how much higher is the 
potential at B than that at A ; the reading B-C shows the amount 
by which the potential at C is higher than that at B. Hence, 
the sum of the two readings indicates the difference of potential 
or the voltage between C and A. Consequently, the sum of all 
the readings around the closed circuit indicates the difference 
of potential between A and A, which difference is evidently 
zero, no matter by which closed path the original point has been 
reached. The reader may again resort to analogues in order to 



Chap. II] DIRECT-CURRENT CIRCUITS 19 

see this law more clearly. For instance, the potentials at the 
joints may be likened to temperatures, and the voltmeter readings 
to differences of temperature. Or the potentials may be com- 
pared to absolute pressures in a network of pipes, and the voltages 
e t to the differences of pressure. Again, the potentials of the 
points A, B, C, etc., are analogous to the altitudes of certain 
points, say above the sea level, while the voltages correspond to 
their relative elevations. In all such cases the sum of the differ- 
ences around a closed path is equal to zero. 

Equation (27) is usually written in a somewhat different 
form, because the values of e t are usually not known, so that it 
is desirable to express them through the given electromotive 
forces and the resistances of the conductors. The general expres- 
sion (4) of Ohm's law holds for each conductor in the network. 
Write these expressions for all the conductors along a closed 
path, and add them together, term by term. The sum of the 
e/s is equal to zero, according to eq. (27), so that the result is 

2e* = Sir (28) 

This form of eq. (27) is known as KirchhorFs second law. In 
this equation a certain direction of currents and voltages must be 
assumed as positive. Let, for instance, in the circuit ABCDEFA 
the clockwise direction be taken as positive; that is, all the cur- 
rents flowing clockwise are to be considered positive, and also 
all the e.m.fs. which tend to produce currents in the clockwise 
direction. Let e\ = 70 volts, and e 2 = 50 volts, and let the 
resistances of the conductors be 2, 3, 5, 4, 8 and 6 ohms respec- 
tively. Let all the currents be known except that in DE, and let 
them be 10, 15, 15, 3 and 5 amp. respectively, the directions being 
those shown in the figure. Denote the unknown current in DE 
by x, and assume it to flow in the clockwise direction. Equation 
(28) then becomes 

70-50=10X2 + 15X3-15X5 + 4z + 3X8 + 5X6, 

from which x = — 6 amp. In other words, the current in DE is 
equal to 6 amp. and is flowing counter-clockwise. 

For a given network of conductors the number of equations 
of the form (26) is equal to the number of junction points less one, 
because the equation for the last point can be obtained by com- 
bining the other equations. The number of equations of the 



20 



THE ELECTRIC CIRCUIT 



[Art. 7 



form (28) is equal to the number of independent closed paths in 
the network. The total number of equations of both kinds is 
just equal to the number of unknown currents, so that these 
currents can be determined by solving the simultaneous equations. 

Prob. 1. A constant e.m.f. of 110 volts is maintained at the generat- 
ing station, and power is transmitted through a line having a resistance 
of 0.5 ohm to two devices in parallel, viz., a resistor of 10 ohms, and a 
motor the internal resistance of which is 5 ohms. Calculate the line 
current (a) when the motor armature is blocked, and (b) when it revolves 
at such a speed that the counter-e.m.f. is 90 volts. 

Ans. 28.7 amp.; 13.05 amp. 




Ba. 



Fig. 5. An unbalanced Wheatstone bridge. 



Prob. 2. Write the equations for the six unknown currents in a 
Wheatstone bridge (Fig. 5) when it is not balanced. 

Ans. i b = U + U\ U = i\ + i g ) is = U + i g ; Wb + itf\ + Ur 2 = e; 
i\T x — i> 3 — igfg = 0; i 2 r 2 — i>4 + i a r a = 0. 

Prob. 3. Show that the preceding six equations are reduced to three 
when the galvanometer circuit is open. 




Fig. 6. A leaky electric circuit containing a counter-e.m.f.; this is an 
analogue to the magnetic circuit of a loaded electric machine. 

Prob. 4. Two sources of e.m.f., ei and e 2 (Fig. 6), are connected to act 
against each other, e x being larger than e 2 . The internal resistances of 
these sources are n and r 2 respectively; the main external resistance is 
2 r. The insulation between the terminals of the sources of e.m.f. is 



Chap. II.] DIRECT-CURRENT CIRCUITS 21 

imperfect, and the leakage conductances are represented by gn and gn 
respectively. Write Kirchhoff's equations for the unknown currents 
(a) when there is no leakage; (b) when g i2 = 0; (c) when gn = 0; and 
(d) when both leakages are present. 1 

Prob. 5. A telegraph line with ground return has a resistance of r' 
ohms per kilometer, and a leakage conductance to the ground of g' mhos 
per kilometer. 2 When the line current at the sending station is I, what 
is its value at a distance of x km. from the station? Solution: Consider 
an infinitesimal length dx of the line at a distance x from the sending 
station, and let the voltage to the ground at this point be e. Let the 
current be i, then the leakage current is di, and we have, according to 
Ohm's law, —di = eg' dx, where g' dx is the leakage conductance through 
the element dx of the line. For the element dx of the line itself, Ohm's 
law gives -de = ir dx. The sign minus is necessary in both equations, 
because, according to the physical conditions of the problem, both i and e 
decrease with increasing x. Substituting the value of e from the first 
equation into the second, gives d 2 i/dx 2 = r'g'i, or i is such a function of x 
that its second derivative is proportional to the function itself. The 
only such function known is the exponential function, so that i = Ce~~ ai , 
where C and a are the constants of integration, to be determined from 
the conditions of the problem. We see at once that the factor C = /, 
because i must be equal to / when x = 0. The sign minus in the exponent 
is necessary because i decreases as x increases. Substituting the expo-, 
nential expression for i into the differential equation gives a = vr'g'. 

Prob. 6. Referring to the preceding problem, at what distance from 
the sending station does the current drop to one half of its original value, 
if the resistance of the line is 7 ohms per km. and the insulation resistance 
is 1.2 megohms per km.? Ans. 287 km. 

1 The electric circuit shown in Fig. 6 is of importance because it serves 
as a good analogue to the magnetic circuit in a loaded machine. The electro- 
motive forces Ci and e 2 correspond to the magnetomotive forces of the field 
and the armature respectively, the reluctances of the parts of the main path 
being represented by 2 r, r\ and r 2 , while the leakage permeances correspond 
to gn and gi 2 . See the author's Magnetic Circuit, the latter part of Art. 40, 
and problem 13. 

2 Primed symbols are used in this book and in the Magnetic Circuit where 
quantities refer to unit length. 



CHAPTER III 

CONDUCTORS OF VARIABLE CROSS-SECTION 1 

8. Current Density and Voltage Gradient at a Point. When 
the cross-section of a conductor varies along its length (Fig. 7), 
the voltage drop per unit length and the current density are also 
variable. In places like MN, where the cross-section of the 
conductor is comparatively small, the resistance per unit length 
is correspondingly large, and vice versa. Consequently, the volt- 
age gradient and the current 'density are also larger at MN than, 
for example, at PQ. Equations (23) and (24) give in this case 
only an average current density and an average voltage gradient 
over the conductor. 




Fig. 7. A conductor of variable cross-section, showing the stream lines and 
equipotential surfaces. 



The lines traversing the diagram (Fig. 7) represent stream 
lines and equipotential surfaces. The stream lines, marked with 
arrowheads, represent the direction of the electric flow, while the 
equipotential surfaces are perpendicular to them, and are the loci 
of points of equal potential. The distribution is analogous to that 

1 This chapter may be omitted if desired, because it is not necessary for 
the understanding of the following chapters on alternating currents. The 
importance of this chapter lies in the fact that the treatment is analogous 
to that of the electrostatic circuit, and therefore it greatly facilitates the study 
of the latter. This chapter may, therefore, be conveniently studied before 
taking up Chapter 14. The treatment is also analogous to that used in the 
author's Magnetic Circuit. 

22 



Chap. Ill] CONDUCTORS OF VARIABLE CROSS-SECTION 23 

which obtains in the flow of heat; the stream lines indicate the 
direction of the flow of heat, while the equipotential surfaces are 
analogous to those of equal temperature. 

In order to understand the meaning of equipotential surfaces, 
let one lead of a voltmeter be applied at one of the terminals of 
the conductor, and let the other lead be moved about inside the 
conductor (assuming this to be possible), marking the points for 
which the deflection of the voltmeter remains the same. All 
the points for which the reading is, let us say, 10 volts form an 
equipotential surface; while all those for which the voltmeter 
reads 11 volts form another equipotential surface, and so on. 
Between two points on the same equipotential surface the volt- 
meter reading is evidently zero. The equipotential surfaces are 
perpendicular to the stream lines, because if there were a com- 
ponent of flow along an equipotential surface there would be an 
ir drop between two points on the same surface, and the voltage 
between these two points could not be zero. 

Stream lines and equipotential surfaces give a clear idea of 
the character of flow of a current in a conductor of irregular shape, 
especially if they are drawn to correspond to equal increments of 
current and voltage. This means that the lines of flow should 
be drawn so as to define tubes of current of equal strength. For 
instance, in Fig. 7 the current included between any two adjacent 
stream lines is supposed to be the same — let us say, equal to one 
ampere. Similarly, the voltage between any two adjacent equi- 
potential surfaces should be the same; for example, one volt. If 
the lines are drawn sufficiently close together, they give complete 
information about the voltage and current relations in the differ- 
ent parts of the conductor, and also show places of high and low 
current density and voltage gradient. 

The true current density at a point is obtained by considering 
an infinitesimal tube of current di and dividing di by the infini- 
tesimal cross-section dA of the tube at the point under consider- 
ation. Then, instead of eq. (23), we have 

U=di/dA (29) 



If, on the other hand, it is desired to express the total current 
through the density, the preceding relation gives 

A UdA, ....... (30) 



i 



24 THE ELECTRIC CIRCUIT [Art. 8 

the integration to be extended over the whole equipotential sur- 
face, U being a function of the position of dA. In other words, 
current is the surface integral of current density. 

The relation between the variable intensity G along the conduc- 
tor, and the total voltage e at its terminals, is no longer expressed 
by the simple relation (24), applicable to the whole conductor. 
Relation (24) must now be written for an infinitesimal length dl 
of a stream line, because G is constant only for an infinitesimal 
length. The definition of G remains the same, namely, G is the 
rate of variation of voltage per unit length of the conductor. 
Thus, denoting by de the voltage between two adjacent equi- 
potential surfaces at a distance dl apart, we have 

G = de/dl, or de = G-dl (31) 

The total voltage e between the terminals of the conductor is 
equal to the sum of these infinitesimal drops, or 



-£ 



G-dl (32) 



Equation (32) is expressed in words by saying that voltage is the 
line integral of electric intensity (or voltage gradient) . 

A clear understanding of relations (31) and (32) is of para- 
mount importance in the study of electrostatic and magnetic 
phenomena. This will be aided by recalling to mind the thermal 
analogy previously used. In the case of the flow of heat, G 
corresponds to the rate of change in temperature per unit length 
of the rod, while e represents the total difference of temperature 
between the ends of the rod. Equation (31) expresses the fact 
that, by taking the rate at a certain point and multiplying it by 
a very short element of the length of the rod, the actual difference 
of temperature between the ends of this element is obtained. 
Thus, for instance, let the drop in temperature at some point 
of the rod be equal to 2.5° C. per meter length. Then the actual 
drop in a very short element, say 0.1 mm., is 2.5 X 0.0001 = 
0.00025° C. The element of length must be small, because by 
supposition the cross-section of the rod* is not constant, and the 
rate of drop is consequently variable. For a short length the vari- 
able quantities can be assumed constant, or, more correctly, aver- 
age values can be used. Equation (32) thus states that the total 
difference of temperature between the ends of the rod is equal to 
the sum (or the integral) of the drops in the very small elements. 



Chap. Ill] CONDUCTORS OF VARIABLE CROSS-SECTION 25 

Similarly, in a pipe of variable cross-section the rate of loss 
of head per unit length is variable, so that it is only possible to 
speak of this rate G at a point. The total loss of pressure, or 
head e, is obtained by summing up the small losses of head in 
infinitesimal elements of the pipe. The loss of pressure for a 
length dl isGdl; the total head e is the integral of this expression 
over the whole length of the pipe. This is expressed mathemati- 
cally by eq. (32). 

Relation (25) between G and U holds true for a non-uniform 
flow as well, because it merely gives a relation between the cause 
G and the effect U at a point, depending only upon the property 
of the material, as expressed by the factor y or p. This relation 
may be also considered as Ohm's law for an infinitesimal cylindri- 
cal conductor of length dl and cross-section dA, namely, 

Gdl = de= (pdlfdA)UdA. 
Canceling dA and dl, relation (25) is obtained. 

Prob. 1. A current of 50 amp. is flowing along a cylindrical con- 
ductor 3 cm. in diameter. The resistivity of the material varies in con- 
centric layers in such a way that the current density is proportional to 
the cube of the distance from the axis. What is the current density at 
the periphery? Ans. 17.7 amp. per sq. cm. 

Prob. 2. A conductor of circular cross-section, 225 cm. long, has the 
form of a truncated cone, the diameters of the two terminal cross-sections 
being 1.2 cm. and 3 cm. respectively. The total drop at a certain current 
is 65 volts. What is the general expression for the voltage gradient G x 
at a distance x from the smaller end? 

Ans. G x /G = [a/ (a + x)] 2 , where a = 150 cm. is the distance from 
the smaller end to the apex of the cone, and G = 0.723 volt per centimeter 
is the voltage gradient at the smaller end. G is determined from eq. 

X225 
a 2 dx/(a + x) 2 . 

Prob. 3. A non-linear irregular conductor, made of homogeneous ma- 
terial, has a current density U and an electric intensity G, varying from 
point to point in magnitude and direction. What is the general expres- 
sion for the power converted into heat? 

Ans. According to eq. (25a), 



= fG-Udv = - Cu 2 dv = 7 fG 2 dv t 



(33) 



where dv is the element of volume to which G and U refer, and the inte- 
gration is extended over the whole volume of the conductor. The 
volume dv must be taken as a cylinder or parallelopiped, the length of 
which is in the direction of flow of the current, the cross-section being 
perpendicular to this flow. 



26 



THE ELECTRIC CIRCUIT 



[Art. 9 



9. The Radial Flow of Current. The solution of problems 
involving a non-uniform flow of current usually requires con- 
siderable facility in the use of higher 
mathematics beyond ordinary calculus. 
An exception to this statement is the 
simple case of radial flow (Fig. 8) be- 
tween two concentric electrodes, cyl- 
indrical or spherical. The following 
exercises give an opportunity for prac- 
tice in the solution of problems of this 
kind. They serve to illustrate the 
concepts of current density and voltage 
gradient, and to prepare the student's 
mind for the solution of certain prob- 
lems on concentric cables, involving 
the dielectric and magnetic circuits. 




Fig. 8. Flow of current be- 
tween two concentric elec- 
trodes. 



Prob. 1. Calculate the resistance of a cylindrical layer of mercury 
MM (Fig. 8) of height h = 5 cm., between two concentric cylindrical 
terminals Ti and T 2 , the radii of the contact surfaces being a = 10 cm. 
and b = 18 cm. The resistivity of mercury is 95 microhms per cubic cen- 
timeter. Solution: Take an infinitesimal layer of the mercury, between 
the radii x and x + dx; the resistance of this layer is p • dx / (2 irxh) . 
The resistances of all the infinitesimal concentric layers are in series; 
therefore, r is obtained by integrating the foregoing expression between 
the limits a and b, the result being 



r= [ P /(27r/i)].Ln(6/a). 



(34) 



Ans. r = 1.775 microhms. 

Prob. 2. In the preceding problem, when a current of 10,000 amp. 
flows through the mercury, what is the amount of heat generated per 
second per cubic centimeter of mercury, at both electrodes? Solution: 
The current density at the inner electrode is 10,000/(2 w X 10 X 5) = 
31.8 amp. per square centimeter. According to eq. (25a), the loss of 
power is (31. 8) 2 X 95 X 10~ 6 = 0.0958 watt per cubic centimeter. The 
heat loss at the outer electrode is 0.096 X (10/18) 2 = 0.0295 watt per 
cubic centimeter. 

Prob. 3. What is the curve of electric intensity G as a function of x 
in the preceding two problems, and what are the limiting values of G? 

Ans. An equilateral hyperbola; Gi = 3.02, G% = 1.677 millivolts per 
centimeter length. 

Prob. 4. A lead-covered cable, consisting of a solid circular conductor 
of A square millimeters in cross-section, is insulated with a layer of 
rubber c mm. thick, between the conductor and the sheathing. What is 



Chap. Ill] CONDUCTORS OF VARIABLE CROSS-SECTION 27 

the insulation resistance of I kilometers of such a cable, if the resistivity 
of rubber is p megohms per centimeter cube? 

Ans. [ P X 10- 5 /(2 ttQ] • Ln (1 + 1.772 c/ VI) megohms, according to 
eq. (34). 

Prob. 5. Show that by doubling the thickness of the insulation in 
the preceding problem, the insulation resistance is increased less than 
twice. 

Prob. 6. A current is flowing through a hemispherical shell of metal 
along radial lines. Express the resistance of the shell as a function of its 
radii a and b, and the conductivity y of the material. 

Ans. (b — a) / (2 iryab) . 

Prob. 7. Apply the method of superposition and the result obtained 
in Arts. 60 and 63, to the calculation of the resistance of an unlimited 
conducting medium between two parallel cylindrical terminals. Such a 
case obtains, for instance, when a load resistor consists of two vertical 
pipes in a pond, the pipes being used as the terminals, the current 
flowing through the water. 

10. The Resistance and Conductance of Irregular Paths. 

Let a conductor of irregular shape (Fig. 7) be connected to a 
source of constant voltage e. The power converted into heat in 
the conductor is e 2 /r, where r is the resistance of the conductor. 
This resistance depends upon the distribution of the current in 
the body of the conductor. The general law, demonstrated by 
all experiments, is that the distribution of the current is such as 
to make the dissipated energy a maximum. Since by supposi- 
tion e is constant (unlimited supply), the resistance r must be a 
minimum. 

Let now the same conductor be connected to a source of 
constant current — for instance, an arc-light machine. The dis- 
tribution of the current in the conductor is such as to effect its 
passage with a minimum expenditure of energy, that is, minimum 
voltage at the terminals, or minimum i 2 r. This again means 
that the resistance r is a minimum. The student is advised to 
consider similar cases in the flow of heat or of a fluid, in order to 
make the matter perfectly clear to himself. 

The general law of nature — that of minimum effort or mini- 
mum resistance — applies in all such cases, and is used in the cal- 
culation of the resistance of conductors of irregular form. The 
conductor is divided into small parts by means of stream lines 
and equipotential surfaces as shown in Fig. 7, drawing them to the 
best of one's judgment. These small cells are nearly cylindrical 
in form, so that their resistances or conductances are easily esti- 



28 THE ELECTRIC CIRCUIT [Abt. 11 

mated by using their mean lengths and average cross-sections. 
The resistance of the whole conductor is found by properly com- 
bining the resistances of these cells in series, and the conductances 
of the filaments thus obtained in parallel. Then the assumed 
shapes of the stream lines and of the equipotential surfaces are 
somewhat modified, and the resistance is calculated again, and 
so on. Thus, by successive trials, the minimum resistance, or the 
maximum conductance, of the given conductor is found, and this 
is the true value of resistance or conductance, as the case may 
be. The lines corresponding to this minimum give the true distri- 
bution of currents and voltages within the conductor. 

The work of the trials is made more systematic by following 
a procedure suggested by Lord Rayleigh, and further developed 
by Dr. Lehmann. This method is described in detail in Art. 54 
below, in application to the electrostatic field, and also in Art. 41 
of the author's Magnetic Circuit, in application to the magnetic 
field. The student will have no difficulty in applying the method 
to an electro-conducting circuit. The best way to make it clear 
to one's self is actually to draw a conductor of irregular shape 
(in two dimensions for the sake of simplicity) and to calculate its 
resistance in the above-mentioned manner. 1 

11. The Law of Current Refraction. The method out- 
lined above for the mapping out of stream lines and equipoten- 
tial surfaces applies only in a homogeneous conductor. When a 
current passes from one substance to another (Fig. 9), the stream 
lines suddenly change their direction at the dividing surface AB 
between the media, and in so doing they obey the law of cur- 
rent refraction, which is 

tan0i/tan0 2 = 7i/Y2.'. ..... (35) 

Here 0i and d 2 are the angles of incidence and refraction, while 
7i and 72 are the respective conductivities of the two media. 
This equation shows that the lower the conductivity of a sub- 
stance, the more nearly do the stream lines approach the direc- 

1 In two-dimensional problems of this kind, the properties of conjugate 
functions may be used when the geometric forms involved can be expressed 
by analytic equations. However, the purely mathematical difficulties are 
such as to make this method applicable only in a comparatively few simple 
cases. See J. C. Maxwell, Electricity and Magnetism, Vol. 1, p. 284; J, J. 
Thomson, Recent Researches in Electricity and Magnetism, chap. 3; Horace 
Lamb, Hydrodynamics, chap. 4. 



Chap. Ill] CONDUCTORS OF VARIABLE CROSS-SECTION 



29 



tion of the normal iVjiV 2 at the dividing surface. In this way, 
the path between two given points is shortened in the medium 
of lower, and is lengthened in that of higher conductivity, by such 
an amount in each case that the total conductance of the composite 
conductor is larger with refraction than without it. Hence, the 
existence of refraction is a necessary consequence of the general 
law of least resistance. 



Medium 
of low conductivity 




Medium 1, 
of high conductivity 



Fig. 9. The refraction of a current, or of a flux. 



To deduce eq. (35), consider a tube of current between the 
equipotential surfaces ab and cd, and let the width of the path in 
the direction perpendicular to the plane of the paper be one 
centimeter. Let XJ\ and Ui be the current densities in the tube, 
and let G\ and G 2 be the corresponding voltage gradients. Two 
conditions must be satisfied, namely, (1) the total current through 
cd is equal to that through ab, and (2) the voltage drop along ac 
is the same as that along bd. These conditions are expressed by 
the equations 

XJi • ab = U 2 • cd 
and 

G\ • bd = Go • ac. 

Dividing the first equation by the second and rearranging the 
terms gives 

Vi/Gi = U2/G2 . 

bd/ab ac/cd 



30 THE ELECTRIC CIRCUIT [Art. 11 

But, according to eq. (25),_J7i/(ti = 71, and U2/G2 = 72. From 
Fig. 9, bd/ab = tan 6 h and ac/cd = tan 2 . By substituting these 
values in the preceding equation, relation (35) is obtained. 

Thus, in mapping out an electro-conducting circuit in two 
media, the stream lines must be so drawn as to satisfy eq. (35), 
and the conductance must be a maximum for the combination, 
and not for each part separately. A similar law applies to elec- 
trostatic and magnetic fluxes (see Art. 55 below, and Art. 41a of 
the author's Magnetic Circuit). 

Prob. 1. Make clear to yourself the reason why the refraction of 
light follows a sine law, while in the case of the electric current it is a 
law of tangents. 

Prob. 2. Show that total refraction is impossible for an electric cur- 
rent. 

Prob. 3. Draw a set of curves giving values of 0i for different values 
of 02 when the ratio of conductivities is 1, 2, 10 and 100. 



CHAPTER IV 

REPRESENTATION OF ALTERNATING CURRENTS AND 
VOLTAGES BY SINE-WAVES AND BY VECTORS 

12. Sinusoidal Voltages and Currents. A large proportion 
of the electric power used for lighting, industrial purposes, and 
traction is generated in the form of alternating currents. Some 
of the advantages of the alternating current over the direct cur- 
rent are: (1) Alternating-current power can be easily converted 
into power at a higher or at a lower voltage, thus making possible 
the transmission of power over long distances; (2) the genera- 
tion of alternating currents is simpler than that of direct currents, 
the latter requiring a commutator, 1 which needs constant atten- 
tion in operation; and (3), by combining two or three alternating- 
current circuits into a polyphase system it is possible to convert 
electric into mechanical power, using motors of simple and rugged 
construction (induction motors and synchronous motors). 

Alternating voltage waves generated by commercial alter- 
nators are more or less irregular in shape, but for most engineer- 
ing calculations it is accurate enough to assume them to vary 
with the time according to the sine law (Fig. 10). This assump- 
tion simplifies the theory and calculations greatly; moreover, 
the results obtained with this assumption are comparable with 
one another, because they all refer to a standard shape of the 
voltage and current curves, instead of a particular form in each 
specific problem. If the curve of a voltage or current differs 
greatly from the sine-wave, it can be resolved into a series of sine- 
waves of different frequencies, so that even then the sine-wave 
remains the fundamental form (see Art. 15 below). Fig. 10 
shows the well-known construction of a sine-wave, the instan- 
taneous values of the current or voltage being represented as 

1 The homopolar machine, which is a direct-current machine without a 
commutator, has not proven, up to the present time, to be commercially suc- 
cessful. 

31 



32 



THE ELECTRIC CIRCUIT 



[Art. 12 



ordinates, against time as abscissa?. Instead of actual time in 
seconds, the curve is sometimes plotted against some other quan- 
tity proportional to time — for instance, fractions of a complete 
cycle. It is sometimes convenient to use as abscissae the angu- 
lar positions of a field pole of the alternator with respect to an 
armature conductor in which the electromotive force under con- 
sideration is induced. 




Fig. 10. An alternating current represented by a sine-wave. 



To construct the curve of an alternating current or voltage, 
draw a circle the radius of which equals the maximum value of 
the wave. Divide the circle into a certain number of equal or un- 
equal parts, such as ab, be, etc., and mark on the axis of abscissae 
points a', b', c', etc., corresponding to the points of division on 
the circle. That is, a'b' is either equal or proportional to ab; 
b'c' is either equal or proportional to be, and so on. In general, 
an abscissa such as a'e' represents, to a certain selected scale, the 
central angle u corresponding to the arc ac. The length a'm' 
represents to the same scale an angle of 360 degrees, or the time 
of one complete cycle of the wave. The ordinates of the sine- 
wave are equal to the corresponding ordinates of the circle. For 
example, the point c'" on the curve is obtained by transferring 
the ordinate ee" of the circle to the corresponding abscissa a'e' . 

The name " sine-wave " is derived from the fact that these 
ordinates are proportional to the sines of the abscissae, which 
represent to some scale the central angles of the circle of refer- 
ence. The equation of the curve expresses this property ana- 
lytically. Let the maximum value of the current, which is also 



Chap. IV] SINE-WAVES AND VECTORS 33 

equal to the radius of the circle, be denoted by I m ; we have then 

from the triangle Occ" 

i = I m sin u, (36) 

where the ordinate i = cc" = c'c'" represents the instantaneous 
value of the alternating current, at the moment of time corre- 
sponding to the angle u. The variable angle u is proportional 
to the time, because the radius Oc which generates the sine-wave 
is assumed to revolve at a uniform speed. Let time t be counted 
from the position Oa of this radius, and let T =a'm' be the in- 
terval of time necessary to complete one revolution of the radius, 
or the time of one complete cycle of the alternating wave. When 
t = 0, u = 0; and when t = T, u = 2 ir. Therefore, in general, 

u = 2Trt/T, (37) 

because this expression satisfies the foregoing conditions. Sub- 
stituting this value of u into eq. (36), we obtain 

i = I m sm(2Trt/T) (38) 

For the values of t = 0, \ T, T, | T, etc., i = 0, as one would 
expect, because at these moments the current changes from 
positive to negative values, or vice versa. At t = \ T, f T, f T, 
etc., we have i = zLl m ; at these moments the current reaches 
its positive and negative maxima. Equation (36) is used when 
the sine-wave is plotted against the values of angle as abscissae. 
Equation (38) gives the same curve referred to time as abscissae. 
In practice, the rapidity with which currents and voltages 
alternate is not denoted by the fraction of a second T during 
which a cycle is completed, but, in a more convenient manner, 
by the number of cycles per second. Thus, instead of saying 
that an alternator generates current which completes a cycle 
within eV °f a second, it is customary to say that the frequency 
of the current is 60 cycles per second. Denoting the frequency 
in cycles per second by /, we have 

/ = i/r, '. (39) 

and consequently 

i = / m sin27r/^ (40) 

This is the usual expression for an alternating current having a 
frequency of / periods per second. Analogously, for an alternat- 
ing voltage we have 

e = E n 8ui2irft, (41) 



34 



THE ELECTRIC CIRCUIT 



[Art. 12 



where E m is the maximum instantaneous value, also called the 
amplitude, and e is the instantaneous value of the voltage at the 
time t. 

In numerical calculations, and when drawing sine-waves, the 
values of the ordinates for various values of u or t are obtained 
either graphically, as in Fig. 10, or from a table of sines. For 
approximate calculations, values of sines can be taken from a 
slide-rule. In the problems which follow, the student is advised 
to become familiar with each of the three methods of obtaining 
values of sines. 

In some cases one has to deal with two currents or voltages 
of the same frequency — for instance, in two different parts of the 
same circuit. The two corresponding sine-waves (Fig. 11) usu- 




Fig. 11. Two alternating currents displaced in phase by an angle <f>. 



ally differ in amplitude, and also pass through zero at different 
instants. Thus, in Fig. 11, when the current 1 is at a maximum, 
the current 2 is still growing, and passes through its maximum 
somewhat later. In other words, the current 2 lags behind the 
current 1, or, what is the same, the current 1 leads the cur- 
rent 2. The angle <f> between the zero points (or between the 
maxima) of the two waves is called the angle of phase differ- 
ence, or simply the phase angle. If we regard the two waves 
as being formed by the revolving radii I m f and I m ", then <f> is the 
angle between the radii at any instant. 

When the two waves are of different frequencies, there is no 
constant phase angle between them; but this angle varies peri- 
odically, so that at some intervals of time the two waves are 
nearly in phase, and at others they are nearly in opposition. 
The familiar method of synchronizing two alternators by means 
of incandescent lamps is based on this phenomenon. 



Chap. IV] SINE- WAVES AND VECTORS 35 

Prob. 1. An alternating current fluctuates according to the sine law 
between the values of ±75 amp., making 6000 alternations per minute 
(3000 positive and as many negative ones). Draw a curve of instan- 
taneous values of this current; mark on the axis of abscissae the time t in 
thousandths of a second, the angles u in degrees, and the same angles in 
radians. 

Prob. 2. What is the frequency of the current in the preceding prob- 
lem, in cycles per second? Ans. 50. 

Prob. 3. Plot on the same curve sheet with the curve obtained in 
problem 1 the sine-wave of a current the frequency of which is three times 
as great, and the amplitude, 52 amp. The curve is to be at its maximum 
when the first curve is at a maximum. 

Prob. 4. Supplement the preceding- curves by one, the frequency of 
which is 50 cycles per second, the amplitude 63 amp., and which reaches 
its positive maxima at the same instants in which the first curve passes 
through zero. Show that with these data two distinct curves can be 
drawn. 

Prob. 5. Draw on the same curve-sheet with the preceding curves a 
sine-wave representing a 50-cycle alternating current, the amplitude of 
which is 120 amp., and which lags by 30 degrees with respect to the 
current in problem 1. 

Prob. 6. The current mentioned in problem 1 is generated by a 
12-pole alternator, that in problem 3 by a 14-pole machine. At what 
speeds must these machines be driven in order to give the required 
frequencies? Ans. 500 and 1285 r.p.m. 

Prob. 7. Express the currents given in problems 1 to 5 by equa- 
tions of the form of eq. (36). Ans. i = 75 sin u; i = — 52 sin 3 y,; i = 
±63 cos u; i = 120 sin (u - 30°). 

Prob. 8. The angle u in the answers to the preceding problem is ex- 
pressed in degrees; rewrite the equations so as to have u expressed in 
radians, and in fractions of a cycle. Also represent the currents as func- 
tions of the time t. 

Prob. 9. Express by equations similar to eq. (41) the following sinus- 
oidal voltages of frequency /: (a) Amplitude E m volts, (b) Amplitude 
E m f volts, lagging a degrees with respect to the first curve, (c) Ampli- 
tude E m " volts, leading the second curve by <f> radians, (d) Amplitude 
E m '" volts, lagging one nth of a cycle with respect to the curve (a). 

Prob. 10. The voltages required in the preceding problem are induced 
by four identical alternators, having p poles each, and coupled together. 
By what geometrical angles must the revolving or the stationary parts 
be displaced in order to give the required differences in phase? 

13. Representation of a Sine-wave by a Vector. It is clear 
from the foregoing theory and problems that all sine-wave cur- 
rents or voltages are different from one another in three respects 
only, namely: (1) in amplitude; (2) in frequency; and (3) in 
relative phase position. In most practical cases, all the currents 



36 THE ELECTRIC CIRCUIT [Art. 13 

and voltages entering into a problem are of the same frequency, 
so that they differ from each other solely in their amplitudes and 
phase positions. In such cases it is not necessary to draw sine- 
waves, or even to write their equations ; it is sufficient to indicate 
the radii I m ' and I m " which generate these curves (Fig. 11), in their 
true magnitudes and relative positions. The rotating radius, at 
any instant, gives by its vertical projection to scale the magnitude 
of the alternating current or voltage at that instant. 

The absolute position of the radii is immaterial, because they 
are revolving all the time. It is their relative position which is 
permanent, and which determines the relative position of the 
sine-waves. The moment from which time is counted is arbi- 
trary in most problems; hence, one of the radii can be drawn in 
any desired position. Then, all other radii in the same problem 
are determined by their phase displacement with respect to this 
" reference" radius. 

It must be clearly understood that the foregoing representa- 
tion by vectors is true only when all the vectors are revolving at 
the same speed, that is, only with alternating quantities of the 
same frequency. When currents and voltages of different fre- 
quencies enter into a problem, the angle between the vectors 
varies all the time, and it is necessary to introduce an arbitrary 
zero of time for reference. In general, the graphical method of 
solution is unsuitable for such problems. 

In mathematics and physics, a quantity which has not only a 
magnitude, but also a definite direction in space or in a plane, 
is called a vector. Thus, for instance, in mechanics, force is a 
vector quantity, while volume is not. The radii which repre- 
sent sine- waves have both magnitude and direction in a plane. 
It is proper, therefore, to call them vectors. While the direction 
of the first vector is usually arbitrary, once it is selected, the 
directions of all the other radii become definite, so that with 
this limitation, the radii in alternating-current problems have 
definite directions and may be called vectors. While they must 
be imagined as revolving when generating their respective sine- 
waves, yet they revolve as a system, maintaining their relative 
positions unchanged. The required relations always depend 
upon the relative positions of the radii, so that the fact that they 
are revolving can be altogether disregarded, and the radii consid- 
ered as simple stationary vectors. 



Chap. IV] SINE-WAVES AND VECTORS 37 

Prob. 1. Draw the vectors of the currents in problems 1, 4 and 5 of 
the preceding article in their true magnitudes and relative positions. 

Prob. 2. A single-phase alternator has a terminal voltage of which 
the maximum instantaneous value is equal to 16 kilo volts. The maxi- 
mum value of the current supplied by the machine is 325 amp. The 
character of the load is such that the current wave lags behind the voltage 
wave by an angle of 37 degrees. Assuming both the voltage and the 
current to vary according to the sine law, represent the foregoing condi- 
tions by two vectors. 

Prob. 3. Draw a vector diagram showing the phase (star) voltages and 
currents of a 25-cycle three-phase system (Fig. 36), the amplitude of 
each voltage being 7235 volts, and each displaced in phase by 120 degrees 
with respect to the other two voltages. The current in the first phase 
is 30 amp., and lags behind the corresponding phase voltage by | of a 
cycle. The current in the second phase is 47 amp., and leads its voltage 
by 18 degrees. The current in the third phase is 72 amp., and lags behind 
the corresponding phase voltage by 0.0C4 of a second. 

Note: In the foregoing three problems the student is supposed to 
draw the vectors equal in length to the amplitudes of the alternating 
waves. In practice, it is customary to draw vectors equal in length to 
the effective values of voltages and currents, and not to their amplitudes. 
For sine-waves the effective value is equal to the amplitude divided by 
V2 (see Chapter 5). The difference is not important for our present 
purposes. The use of effective values would merely change the arbi- 
trary scale to which the vectors are drawn. 

14. Addition and Subtraction of Vectors. There are many 
practical problems in which alternating currents or voltages have 
to be added, or subtracted one from another. For instance, 
when two or more alternators are working in parallel, the total 
current delivered to the station bus-bars is equal to the sum of 
the currents supplied by each machine. Or, to find the voltage 
at the receiving end of a transmission line, the voltage drop in 
the line is subtracted from the generator voltage. When the com- 
ponent quantities vary according to the sine law and are all of one 
frequency, the resultant quantity is also a sine curve of the same 
frequency. This curve may be found (a) graphically, by adding 
the component curves point by point; (b) analytically, by adding 
their equations; or (c) by adding the vectors of these curves. 

It must first be proved that the sum of two sine-waves of 
one frequency is also a sine-wave of the same frequency. Let 
the two currents to be added be represented by the equations 

i = I m sin (u + <}>) ) 
i' = I m ' sin (" + «') [' •••:■•- M 



38 THE ELECTRIC CIRCUIT [Art. 14 

where u = 2 irft is the variable time angle, and and 0' are two 
constant angles characterizing the relative phase positions of the 
two waves with respect to some reference wave I m " sin u. The 
phase displacement between i and %' is <$>' — 0. Expanding the 
foregoing sines of the sum of two angles, and adding the two 
equations, member for member, we obtain 

ieq = i + i* = (I m cos + I m ' cos 0') sin u 

+ (I m sin + I m ' sin 0') cos u, . . . . (43) 

where the constant coefficients of sin u and cos u are grouped 
together. The subscript eq stands for "equivalent." This ex- 
pression is of the form i eq = A sin u + B cos u, where J. and B 
are constants. No matter what values A and B may have, the 
right hand side of this equation is reducible to the form 

Ieq = IeqmSin(u + (j) eq ) (44) 

Assuming this equation to be true, we equate the right-hand sides 
of eqs. (43) and (44), and expand sin (u + 4> eq ). Equating the 
coefficients of sin u and cos u, we get 



Ieqm COS (j> e q = I m COS + I J COS 0'j 

l eqm sin <j> eq = I m sin + I m ' sin 4/. 



(45) 



These are two simultaneous equations with I eqm and <j> eq as the 
unknown quantities. Squaring and adding these equations, we 
obtain 

Ieqm 2 = (I m sin + IJ sin $')* + (I m cos + 7 W ' sin r ) 2 . (46) 

Dividing the second equation by the first gives 

tan eg = (I m sin + I J sin ({> f )/(I m cos + /„/ cos 0'). (47) 

No matter what values I m , I m r , <t> and 0' may have, the values 
of I eqm and <f>eq determined from these equations are real. In 
other words, it is always possible to represent eq. (43) in the 
form of eq. (44). This proves the proposition, because we see 
from eq. (44) that i eq is a sine-wave having the same u = 2irft for 
the variable angle, hence the same frequency as the component 
waves. The amplitude and the phase position of this resultant 
wave are determined by eqs. (46) and (47) . 

When two currents or voltages are represented by vectors, 
their sum or difference is also a vector, because, as proved before, 



Chap. IV.] 



SINE-WAVES AND VECTORS 



39 



it is also a sine-wave of the same frequency. The problem 
is to find the vector of the resultant wave, knowing the vectors 
of the component waves in their magnitudes and positions. Any 
ordinate of the resultant wave must be equal to the sum of the 
corresponding ordinates of the component waves. Hence, the 
vector of the resultant wave must satisfy the condition that its 
projection upon the F-axis (Fig. 12) shall be equal to the sum of 
the projections of the component vectors on the same axis. This 
condition must be fulfilled at all instants of time, that is, during 
the rotation of the three vectors. To satisfy this requirement 
the resultant vector must be the diagonal of a parallelogram of 
which the other two vectors are the adjacent sides. 




Fig. 12. Addition of vectors. 



Let OA and OB be the given vectors to be added together. 
From the end B of the vector OB draw a line BC equal and paral- 
lel to OA. Connecting and C gives the resultant vector OC, 
in magnitude and position. It will be seen from the figure that 
the projection of OC upon the F-axis is equal to the sum of the 
projections of OB and BC upon the same axis. But BC is equal 
and parallel to OA, so that the projection of OC on the vertical 
axis is equal to the sum of the projections of the given vectors 
on the same axis. This construction holds true for any instant 
whatever. By drawing AC, the parallelogram OBCA is com- 
pleted, so that the construction is identical with that for finding 
the resultant of two mechanical forces, However, in practical 
applications it is not necessary to complete the parallelogram, 
because the resultant is perfectly determined by the triangle 
OBC. The resultant of two vectors obtained in this way is called 
their geometric sum. 



40 



THE ELECTRIC CIRCUIT 



[Art. 14 



If the triangle were not closed, the condition of equality with 
the sum of the projections of -the given vectors might be satisfied 
for one particular instant of the cycle, but would not be satisfied 
for other instants. Thus, for instance, assuming the line OC 
to be the resultant vector, we see that for the instant shown in 
the sketch the projection of OC upon the axis OY is equal to the 
sum of the projections of OB and BC upon the same axis; but the 
condition is not fulfilled when the vectors rotate. 

The rule for subtraction of vectors follows immediately from 
the preceding rule, because to subtract a vector means to add a 
vector with the opposite sign. Thus, let it be 
required to subtract the vector OA from OB 
(Fig. 13); this may mean, for instance, the 
subtraction of the voltage wave represented 
by OA from that represented by OB. From 
the end B of OB draw vector BC equal and 
opposite to OA. The resultant, OC, represents 
the difference of the two given vectors, in 
direction and magnitude, and thus determines 
the sine-wave of the resultant voltage. If it 
were required to sub tract. OB from OA, it 
would be necessary to draw AC equal and 
opposite to OB, thus obtaining the resultant 
OC, equal and opposite to the former resultant 
OC. This is in accord with the general alge- 
Fig. 13. Subtraction braic rule that A - B = -(B - A). 

of vectors. The prece diiTg results with regard to the 

addition and subtraction of vectors are summed up in the follow- 
ing rule : Relations which are true algebraically for instantaneous 
values of sinusoidal currents and voltages, hold true geometrically 
for the vectors of these quantities. It is customary to provide 
vectors of currents with triangular arrows, as in Fig. 12; vectors 
of voltages are usually distinguished by pointed arrows, as in 
Fig. 13. This distinction enables one to see directly from the 
diagram whether a vector represents a current or a voltage, with- 
out reference to the text. 




Prob. 1. The currents generated by two alternators in parallel are 
75 and 120 amp. respectively, the second current lagging behind the first 
by 30 degrees. Determine the magnitude and the relative phase position 



Chap. IV] SINE-WAVES AND VECTORS 41 

of the resultant line current by three methods: (a) point by point; 
(b) analytically; (c) by means of vectors. 

Ans. 188.8 amp., lagging by 18° 32' behind the first current. 

Prob. 2. Solve the preceding problem without the use of eqs. (46) 
and (47), simply by means of the theorem proved above, that the sum 
or the difference of two sine- waves is also a sine-wave. Solution: 

I eq sin {u + <t>eq) = 75 sin u + 120 sin (u - 30°). 
This equation is true for any instant, or for any value of u. It contains 
two unknown quantities, the amplitude and the phase position of the 
resultant curve. It is necessary, therefore, to apply this equation to two 
particular moments of time, in order to obtain two equations with two 
unknown quantities. It is most convenient in this particular case to 
choose u = 7I-/2 and u = 0. Substituting these values, two equations 
with two unknown quantities are obtained. This method is preferable 
in the solution of practical problems, because it is not necessary to 
remember eqs. (46) and (47), and also because the two values of u can 
be selected so as to give the simplest equations. 

Prob. 3. Two alternators, with the same number of poles, are coupled 
together so as to give voltages differing in phase b}^ 27 degrees, the voltage 
of the second machine leading that of the first. The first alternator 
generates a voltage the amplitude of which is 2300 volts, the second 
1800 volts. The two machines are connected electrically in series. Find 
graphically the vector of the resultant voltage in its magnitude and phase 
position. Find also the vector of the resultant voltage when the termi- 
nals of one of the machines are reversed. Ans. (1) 3988 volts, lead- 
ing the first by 11° 49' ; (2) 1074 volts, lagging behind the first by 49° 32'. 

Prob. 4. An alternator, the terminal voltage of which is 6600, supplies 
its load through a transmission line. The conditions are such that the 
current lags behind the generator voltage by an angle of 35 degrees. 
The voltage drop in the line is 540 volts, leading the current in phase by 
an angle of 67 degrees. Find the receiver voltage by subtracting the 
voltage drop in the line from the generator voltage (geometrically) ; also 
determine the phase displacement between the receiver voltage and the 
current. Ans. 6149 volts; 32° 20'. 

15. Non-sinusoidal Currents and Voltages. When a current 
or voltage wave differs considerably from the pure sine form, 
it is often convenient to represent it as the result of a superposi- 
tion of sine-waves of different frequencies (Fig. 14). No mat- 
ter how complicated a periodic wave may be, it can always be 
so represented with sufficient accuracy, by properly selecting the 
amplitudes and the phase relations of the component sine-waves, 
or harmonics, as they are called. Theoretically, an infinite num- 
ber of sine-waves is necessary in order to represent any given 
irregular wave exactly. In practice, however, a limited number 
of harmonics is sufficient. 



42 



THE ELECTRIC CIRCUIT 



[Art. 15 







Chap. IV] SINE-WAVES AND VECTORS 43 

If the frequency of the given irregular wave is/, the frequencies 
of the harmonics are /, 2 /, 3 /, and so on. When, however, the 
given wave is symmetrical, that is, when the part above the axis 
of abscissae is identical with that below, all even harmonics (2/, 
4/, etc.) drop out, and the wave consists only of the fundamental 
wave of frequency /, and the odd harmonics (3/, 5/, etc.). The 
student can easily convince himself of the truth of this statement 
by taking a fundamental sine-wave and adding to it a second 
harmonic and a third harmonic. In the first case the resultant 
wave will be unsymmetrical ; in the second, symmetrical. Nearly 
all of the waves encountered in practice are symmetrical. 

Let the fundamental wave be represented by the equation 
V\ = C\ sin (u — ai), the third harmonic by the equation t/ 3 = 
C 3 sin 3 (u — q- 3 ), etc. The meaning of d, C 3 , etc., and of a h 
a$, etc., is clear from Fig. 14; is an arbitrary origin from which 
the angles are measured. The ordinates of the given composite 
symmetrical wave are represented by the equation 

y = Ci sin (u — «i) + C 3 sin 3 {u — a 3 ) + C b sin 5 (u — a b ) 

+ etc (48) 

This expression is known as the Fourier series, and is of great 
importance in mathematical physics. 

In practice, the problem which presents itself is usually that of 
analysis; that is to say, it is often required to analyze or resolve 
a given irregular wave into its harmonics. In other words, know- 
ing y, one is asked to determine the values of C and a for one or 
more harmonics. This is a purely mathematical problem, and is 
not treated here, because the solution will be found in numerous 
textbooks, handbooks and magazine articles. 1 There are also 
mechanical wave analyzers on the market, by means of which 
any desired harmonic may be separated by tracing the given 
curve with a stylus, in a manner similar to the way in which a 
planimeter is used. 

It is of importance for an electrical engineer to train his eye 
in the discernment of prominent harmonics, without mathematical 
analysis. This training is afforded by exercises in wave synthesis, 
that is, in combining various assumed harmonics into irregular 
waves. 

1 See, for instance, the author's Experimental Electrical Engineering, Vol. 2, 
p. 222. 



44 THE ELECTRIC CIRCUIT [Art. 15 

Take first a fundamental sine-wave and a third harmonic of 
a reasonable magnitude, say between 15 and 30 per cent of the 
fundamental. Combine these waves into one, with different rela- 
tive phase positions of the fundamental and the third harmonic. 
In this way a flat wave, a peaked wave and a one-sided "humped" 
wave will be obtained. Then change the magnitude of the third 
harmonic and construct similar waves, in order to see the influ- 
ence of this factor. After that, plot similar curves for the funda- 
mental wave with a fifth harmonic, a seventh harmonic, and so 
on. Finally, combine the fundamental wave with the third and 
the fifth harmonics simultaneously, and so on. After some prac- 
tice, the eye will easily discern prominent harmonics in a given 
irregular wave. Numerous oscillograms of irregular waves will 
be found in many current periodicals and in the transactions of 
the various electrical societies. Read in this connection Art. 30 
of the Magnetic Circuit. 

Prob. 1. Draw two or three sets of curves suggested in the preceding 
paragraph; each set must comprise about six curves and each curve 
must have a harmonic with a different phase position. 

Prob. 2. Devise a simple apparatus by means of which harmonics 
can be combined mechanically, and the resultant waves observed, with- 
out actually plotting curves point by point. 

Prob. 3. Analyze a given irregular wave into its harmonics, using the 
method given in the reference above, or any other method found in the 
literature on the subject. 



CHAPTER V 
POWER IN ALTERNATING-CURRENT CIRCUITS 

16. Power when Current and Voltage are in Phase. Let 

a resistance r be connected across the terminals of an alternator, 
the voltage at the terminals varying according to the sine law. 
The current through the resistance also varies according to the 
sine law, because Ohm's law holds true for any moment of time, 
so that the curve of the current is in phase with that of the voltage. 
If the equation of the voltage wave is e = E m sin u, the equation 
of the current is i = (E m /r) • sin u. Graphically, the current and 
the voltage are represented by two vectors of different lengths, but 
in the same direction — for instance, like OC and OD in Fig. 13. 

Divide the time T of one cycle into a large number of small 
intervals At. Then the amount of energy delivered to the resist- 
ance r and converted into joulean heat during one of such inter- 
vals varies with the time position of the interval in the cycle, 
in other words, with the instantaneous values of the voltage and 
the current. This energy is practically equal to zero when the 
current and the voltage have values near zero, and it reaches a 
maximum with them. However, the dissipated energy, being in 
the nature of a frictional loss, never becomes negative, because 
whether the current flows in one direction, or in the other, the 
heat liberated, e • i • At = i 2 r • At, is always positive. 

Since the voltage and the current vary with the time, the rate 
of liberation of energy, or the instantaneous power, is also variable. 
The expression P = ei = i 2 r represents the instantaneous power 
as with direct current. If e and i remained constant for one 
second, the energy liberated would be equal to i 2 r. As a matter 
of fact, e and i may be considered constant only during the infini- 
tesimal element of time dt, so that the energy liberated during the 
time dt is i 2 r • dt. Nevertheless, it is proper to say that at the in- 
stant under consideration the energy is liberated at a rate equal to 
i 2 r per second, because (i 2 r • dt)/dt = i 2 r. This is analogous to the 

45 



46 THE ELECTRIC CIRCUIT [Art. 16 

way in which we speak of the instantaneous speed of a body during 
a period of acceleration or retardation. The speed varies from 
instant to instant, so that to say that the speed is v at a certain 
instant merely means that, if the body continued to move at this 
velocity for one second, it would cover a space equal to v. In the 
same sense, the instantaneous power indicates the amount of 
energy which would be developed per second, if the current and 
the voltage suddenly became constant. 

The total energy liberated in the form of heat during one 
complete cycle is 



W 



= / i 2 r-dt = r J i 2 - dt (49) 



When no local e.m.fs. are present, the same energy is repre- 
sented by the expression 

W = f 1 ei-dt. .' (50) 

When there are local e.m.fs. in the part of the circuit under 
consideration, the total energy communicated to it during an 
interval of time is different from that dissipated as heat (Art. 4). 
According to eq. (19), we have 

e t i*dt+ I eii-dt = rl i 2 * dt. . . (51) 

o Jo J Jo 

Suppose, for example, that e* is the counter-e.m.f. of a motor 
in the circuit, "and therefore nearly in phase opposition to e t . Then 
the i 2 r loss on the right-hand side of the equation is the difference 
between the energy supplied to the circuit and that converted into 
mechanical work in the motor. 

The foregoing equations are true whether the current and the 
voltage vary according to the sine law or not. If they are sinus- 
oidal, the integration can be easily performed, and the energy per 
cycle evaluated by the following method. Let the current be 
represented as before by % = l m sin u. Substituting this value 
into eq. (49), we have 

W = I m 2 r f T sin 2 udt (52) 

This expression is easily integrated by using the substitution 
sin 2 u = \ (1 — cos 2 m). Or it may be evaluated by observing 
that its value remains the same if a cosine is substituted for the 



Chap. V] POWER IN ALTERNATING-CURRENT CIRCUITS 47 

sine. This is because the limits of integration are u = and 
u = 2 7r, and in summing up sines or cosines through 2 w we take 
the same quantities, only in a different order. Hence we may 
write 

XT 
'cos 2 udt (52a) 

Adding the two expressions term by term, and remembering that 
sin 2 u + cos 2 u = 1 , we get 

2 W = Ur C dt = IJr • T, 

or the energy converted into heat during one cycle is 

W = \IJr-T (53) 

When there are no local e.m.fs. and the current is in phase 
with the voltage, we haye E m = I m r, so that from eq. (53), and by 
analogy to eq. (18), we have 

W = iI m E m -T . (54) 

and 

W = i{E m 2 /r).T (55) 

The student must clearly understand that the phase relation 
between the current and the voltage is of no consequence in eq. 
(53), while eqs. (54) and (55) hold true only when the current is 
in phase with the voltage. Or else, E m in these latter expressions 
may be said to refer to that component of the total terminal 
voltage which is used up in Ir drop. 

Prob. 1. A sine-wave alternating current, which fluctuates between 
±75 amp., flows through a resistance of 10 ohms. Plot curves of instan- 
taneous values of the voltage and power. 

Ans. E m = 750 volts; max. power = 56.25 kw. 
Prob. 2. Determine the total energy liberated per cycle in the pre- 
ceding problem, by integrating graphically the curve of power. 

Ans. 562.5 joules (watt-seconds). 
Prob. 3. Prove analytically that the curve of power obtained in 
problem 1 is a sine- wave of double frequency, tangent to the axis of 
time. Proof: The equation of the curve is P = I m 2 r • sin 2 u. But from 
trigonometry 

cos 2 u = cos 2 u — sin 2 u = 1 — 2 sin 2 u. 
Substituting the value of sin 2 u from this equation into the expression for 
P, we get 

P = \ I m 2 r — \ I m 2 r • cos 2 u. 



The first term is constant, while the second represents a sine-wave of 



48 THE ELECTRIC CIRCUIT [Art. 17 

double frequency, because 2u = 2rr (2f)t. The first term is never 
smaller than the second, so that P is always positive, and the whole curve 
lies above the axis of abscissae. The second term becomes equal to the 
first and P = 0, only when 2 u is a multiple of 2 ■*. At these points the 
curve is tangent to the axis of abscissae. 

Prob. 4. Deduce eq. (53) directly from (52), expressing sin u in terms 
of the cosine of the double angle, as in the preceding problem. Hint: 
From eq. (37), dt = (T/2 -n) du, and the limits of integration are u = and 

u = 2tt. 

17. The Effective Values of Current and Voltage. In prac- 
tice, it is the average rate of delivery or dissipation of energy 
that is of interest, or, in other words, the average value of the 
variable instantaneous power. This is analogous to using in 
calculations the average speed of a machine, when the actual 
speed varies within certain limits. This average power is found 
by dividing the total energy developed during one cycle by the 
period T of the cycle. When the current varies according to 
the sine law, the total energy per cycle converted into heat is 
expressed by eq. (53). Dividing both sides by T, we find that 
the average power 

■Lave — 2 lm T W^V 

It is convenient to use in eq. (56) a new value of the current, 

I = I m /V2 = 0.707 I m , (57) 

instead of I m , because then the expression for the average power 
becomes identical with that in a direct-current circuit, namely, 

Pave = Pr (58) 

Analogously, if we define 

E = E m /V2 = 0.707 E m , ..... (59) 
eqs. (54) and (55) become 

P ave = E.I ....... (60) 

and 

P ave = E*/r, . (61) 

which are perfectly similar to the corresponding expressions in a 
direct-current circuit. 

E and I, as defined above, are called the effective values of the 
alternating voltage and current respectively. We may say that 
by definition the effective value of an alternating (or variable) 
current is equal to such a constant current which, when flowing 



Chap. V] POWER IN ALTERNATING-CURRENT CIRCUITS 49 

through a resistance, dissipates the same average power as the 
actual variable current. 

This definition of an effective value applies to variable cur- 
rents of any form. It is used, for instance, in determining the 
temperature rise of electric railway motors. During the run of a 
car the current fluctuates within wide limits, but the heating of 
the motor windings is nearly the same as would occur with a 
certain constant current, which is called the effective value of 
the actual variable current. The condition for the same average 
i 2 r loss is 

Pr-T 



XT 
i 2 dt, 



where T is the interval of time for which it is desired to obtain 
the effective value. Hence 

P = (1/T) f 1 i 2 dt (62) 

This equation expresses in mathematical language that P is the 
average value of i 2 , over the period of time T. Taking the square 
root of both sides of this equation, we can also define the effective 
value I as the square root of the mean square of the instantaneous 
values. This definition is true for any form of alternating or 
variable current. The effective voltage is defined by a similar 
expression, so that more generally 

y eff 2 = (l/T)£ T y 2 dt, ..... (63) 

where y denotes the instantaneous values of either current or 
voltage. Commercial ammeters and voltmeters, intended for 
use on alternating-current circuits, are always calibrated to indi- 
cate the effective values of current and voltage (see the author's 
Experimental Electrical Engineering, Vol. I, pp. 38, 44 and 47). 

When an irregular wave of current or voltage is given graphi- 
cally, its effective value is found by taking a sufficient number 
of equidistant ordinates (Fig. 15) and replacing the integration 
in eq. (63) by a summation. Let the half-wave be divided into 
k equal parts, and let y , y h . . . y k be the corresponding ordi- 
nates. Then, according to eq. (63), 

Vef f 2 = (1/fc) (i2/o 2 + 2/i 2 + . . . +i^ 2 ). . i (63a) 



50 



THE ELECTRIC CIRCUIT 



[Art. 17 



The larger the number of ordinates, the more accurate is the value 
of y eff determined by this method. The value of y eff 2 may also 
be found by plotting a curve of y 2 against u, as shown in Fig. 15, 
and determining its mean ordinate by means of a planimeter. 

When a large number of effective values must be determined, 
— for instance, from the records obtained by a graphic ammeter 
during several runs of an electric train, — the squaring of ordi- 
nates becomes a tedious process. Some practical methods, by 
means of which the necessity for squaring ordinates is eliminated, 
are described in the next article. 




Fig. 15. The effective and the average ordinates of an irregular half- wave. 



The effective value of a current or a voltage is also called the 
quadratic mean, to distinguish it from the arithmetical mean value 
defined by the familiar equation 

y ave = (l/T)£ T ydt, ...... (64) 

or for a periodic curve 

y ave =(l/>ir)J ydu (65) 

It will be noted that the upper limit of integration is tt and 
not 2 7r. It is evident that for a symmetrical wave the average 
ordinate over a whole cycle is equal to zero. The average value, 
therefore, always refers to a half-wave. 



Chap. V] POWER IN ALTERNATING-CURRENT CIRCUITS 



51 



For the sine-wave 

[jave 



= (VrnM J 



smudu = (2/ir)y m , 



or 



Vave/ym = 2/tt = 0.637 (66) 

The ratio of the effective to the mean ordinate is called the 
form factor, because it gives an idea of the degree to which the 
curve is flat or peaked as compared to the sine-wave. For a 
sine-wave the form factor is 

(y m /V2)/(2y m M = LU (67) 

For a perfectly flat-topped or rectangular wave, the maximum 
value, the effective value and the average value are all the same, 
so that< the form factor is equal to unity. For very peaked waves, 
the influence of the high middle ordinates is more prominent in 
the quadratic mean, so that the effective is considerably higher 
than the mean value, and the form factor is larger than 1.11. 

Another ratio which helps in judging about the shape of a 
curve is the so-called amplitude factor, or the ratio of the maximum 
ordinate to the effective value. The author is not aware that 
either the form factor or the amplitude factor is used to any con- 
siderable extent in practice. 

Prob. 1. An electric heater was tested for power consumption on an 
alternating-current circuit, by having an ammeter in series with it, and 
a voltmeter across its terminals. Both instruments were calibrated to 
indicate effective values. The readings were 110 volts and 5.7 amp. 
Assuming the current and the voltage to have been in phase, which is 
nearly the case, what was the average power consumption of the heater, 
and what was its resistance? Determine also the maximum instan- 
taneous values of the current and the voltage, under the supposition of 
the sine law. Ans. 627 watts; 19.3 ohms; 155.56 volts; 8.06 amp. 



y 2 ir- 



-K7T- 

— T 



Vm 



%y, 



Fig. 16. A stepped curve of current or voltage. 

Prob. 2. Determine the average value, the effective value, the form 
factor and the amplitude factor of the curve shown in Fig. 16. 

Ans. 0.75 y m ; 0.791 y m ; 1.055; 1.264. 



52 



THE ELECTRIC CIRCUIT 



[Art. 18 



Prob. 3. Check some of the values of the form factor and the ampli- 
tude factor given in the table in , the Standard Handbook (see Index under 
" form factor")- This will afford practice in calculating effective values 
of curves when they are given by analytic equations of the form y = f(t), 
using eq. (63). 

Prob. 4. Plot an irregular wave, taken from an available oscillograph 
record, and calculate its average and effective values by the point-by- 
point method, or by using a planimeter. 

18. Some Special Methods for Calculating the Effective 
Value of an Irregular Curve. As is mentioned in the preceding 
article, squaring a large number of ordinates in order to find 
the effective value of a curve is a tedious process, and methods 
are available which sometimes lead to the end more quickly. It 
must be admitted, however, that for one who has to do this 
work only occasionally, the plain point-by-point method described 
above is probably the quickest and the most reliable. 



Given 
durve 




Fig. 17. An irregular curve (Fig. 15) and the equivalent sine-wave, 
plotted in polar coordinates. 



(a) Fleming's Method. The given curve (Fig. 15) is replotted 
in polar coordinates (Fig. 17), so that equal polar angles Cl/k cor- 
respond to equal distances ir/k upon the axis of abscissae. The 
ratio between an abscissa u in Fig. 15 and the corresponding polar 
angle in Fig. 17 is of no consequence; in other words, it makes 
no difference what total central angle O corresponds to the total 
distance -k in Fig. 15. The area of an infinitesimal triangle sub- 
tended by a polar angle dca is J y 2 dco, because y is the base of the 



Chap. V] POWER IN ALTERNATING-CURRENT CIRCUITS 53 

triangle, and y do) is its altitude. Thus, the total area of the curve 
in Fig. 17 is 

S = hfy 2 do) (68) 

But, by the defining equation (63) , the effective value, or the quad- 
ratic mean ordinate, of the same curve in Fig. 15 is found from 
the expression 

V«t = <VO)jrV<k. (69) 

because co is proportional to t. Comparing the preceding two 
equations, we find that 

y cf f = 2S/£l (70) 

Since S is easily evaluated, for instance, by means of a planimeter, 
the effective value is calculated from eq. (70) without squaring 
the ordinates, but simply by replotting the given curve in polar 
coordinates. 1 

When the given curve is a pure sine-wave, the corresponding 
curve in polar coordinates is a circle, provided that the angle O 
is selected equal to t. The student can easily prove this for him- 
self, either graphically or analytically. Let y m be the maximum 
ordinate of the sine-curve; then the area of the circle is S = J Try m 2 , 
and from eq. (70) we find y eff = y m /V2. This is the same value 
as found before by a different method. 

When the given curve is not much different from a pure sine- 
wave, the corresponding polar curve approaches a circle in form 
(always provided that 12 = ir). In such cases it is possible to 
determine the area of the polar curve without a planimeter, by 
drawing a circle of equal area as judged by the eye (Fig. 17). 
The effective value is then the same for the given curve and for 
the sine-wave corresponding to this circle, and is equal to the 
diameter of the circle divided by V2. Such a sine- wave is called 
the equivalent sine- wave. It is often convenient in dealing with 
irregular current and voltage waves to replace them by equivalent 
sine-waves, so as to be able to apply an analytical solution, or 
to construct vectors. 

1 For a more detailed treatment and numerous practical applications, 
see C. O. Mailloux, " Methode de Determination du Courant Constant Pro- 
duisant le meme Echauffement qu'un Courant Variable," in the Transac- 
tions of the International Congress of Applications of Electricity, Turin, 1911. 



54 THE ELECTRIC CIRCUIT [Art. 18 

(b) The Effective Value in Terms of Harmonics. When an 
irregular wave is given in the form of a Fourier series, eq. (48), 
the effective value can be expressed through the amplitudes of 
the harmonics. In order to use the expression for y in the funda- 
mental formula (63), we have to square the Fourier expansion. 
This gives terms of two kinds, namely, squares of harmonics, 
and products of pairs of harmonics. Let the nth and the pth 
harmonics be represented by the expressions 

h n = C n &mn(u — a n ) (71) 

and 

h p = C p sinp(u — a p ) (72) 

Then the right-hand side of eq. (63) will contain the following 
terms : 

(1/T) f h n *dt = (C n 2 /T) f T sitfn(u- a n )dt = \ C n 2 ; (73) 
Jo Jo 

(1/T) f h P 2 dt = (C P 2 /T) f^sin 2 p(u - a p ) dt = } C p 2 ; (74) 
Jo Jo 

2 il/T) f T h n h p dt = 
2 (C n C p /T) f T sin n(u - a n ) sin p(u - a p ) dt = (75) 

The values of the first two integrals are found in precisely the 
same way as that of eq. (52) in Art. 16, that is, on the basis of 
the fact that their values do not change if cosines are substituted 
for the sines. The third integral is identically equal to zero, as 
is shown in problem 3 below. Thus eq. (63) becomes 

Veff 2 = (Ci/V2) 2 + (C3/V2) 2 + etc., . . . (76) 

or the square of the effective value of a complex wave is equal to 
the sum of the squares of the effective values of its harmonics. 

Prob. 1. Plot a complex wave consisting of known harmonics and 
determine its effective value (a) by the method given in the preceding 
article; (b) by the Fleming method; (c) from eq. (76). 

Prob. 2. An irregular wave has a third and a fifth harmonic, the 
amplitudes of which are equal respectively to 12 per cent and 4 per cent 
of that of the first harmonic. Show that the effective ordinate is equal 
to 71.3 per cent of the amplitude of the fundamental wave, and lhat the 
average value depends upon the phase positions of the harmonics. 



Chap. V] POWER IN ALTERNATING-CURRENT CIRCUITS 55 

Prob. 3. Prove that expression (75) is identically equal to zero. 
Proof: According to the familiar formula of trigonometry, sin A sin B = 
\ cos (A — B) — \ cos (A + B), we have sin n{u — a n ) sin p{u — a p ) = 
£ cos [(n — p) u + a] — \ cos [(n + p) u + b], where a and b do not 
contain the variable u. Integrating these cosines leads to terms of the 
form sin [(n — p) u + a] and sin [(n + p) u + &]. Since the limits of 
integration are and 2 tt, and n and p are integers, the values of these 
sines at the upper limit are the same as at the lower limit, and conse- 
quently each of the integrals is equal to zero. 

19. Power when Current and Voltage are out of Phase. 

In a majority of practical alternating-current circuits there is a 
more or less pronounced phase displacement between the current 
and the voltage. This is due to the presence of local electro- 
motive forces, the principal among these being as follows: (a) 
The counter-electromotive forces of motors connected into the cir- 
cuit, (b) The electromotive forces induced by alternating mag- 
netic fluxes in the circuit. These fluxes may be created by the 
current itself, or they may be due to the influence of other circuits 
(self and mutual induction), (c) The electromotive forces due to 
the "elastivity" of the dielectric medium surrounding the circuit 
(electrostatic capacity or permittance). 

The actual workings of these causes are discussed more in 
detail in the following chapters. Here it is sufficient to note that 
there are factors which produce local electromotive forces in 
alternating-current circuits, and that they bring about a phase 
displacement between the voltage and the current. Let OB 
(Fig. 13) be the generator voltage, and let OA represent the sum 
of the various local electromotive forces in the circuit. Sub- 
tracting OA from OB, the net voltage OC is obtained, which is 
just, sufficient to supply the ohmic drop in the circuit. The 
current OD is in phase with this voltage, and is numerically equal 
to OC divided by the total resistance r of the circuit. It will be 
seen that there is a phase displacement <f> between the current and 
the generator voltage OB; it is also clear from the figure that this 
phase displacement is due to the presence of the electromotive 
force BC. 

We shall first calculate the energy supplied by the generator 
during one cycle in the specific case when the phase displacement 
between the current and the voltage is exactly 90 degrees. If 
the current is represented by the equation i = I m sin u, the 
expression for the voltage is e = E m cos u. The instantaneous 



56 THE ELECTRIC CIRCUIT [Art. 19 

power is equal to i • e = I m E m sin u cos u = \ I m E m sin 2 u. Thus, 
the power varies as a sine function of double the generator fre- 
quency; the energy flows now away from, and now toward, 
the generator. The average power for one cycle is therefore zero, 
for the power has as many negative values as it has positive ones. 
Mathematically, this result is represented by the time integral 
of the instantaneous power over a complete cycle. Omitting the 
constant quantities E m and I m , we have 

X2ir T ~]27T 

sin u cos u du = J — cos 2u\ = 0. 

Let now the phase displacement between the current and the 
voltage be less than 90 degrees, and be equal, say, to <£. The 
average power delivered by the alternator is in this case smaller 
than the product EI, and its value must be investigated. The 
vector of the voltage E can be resolved into a component E cos </> 
in phase with the current, and another component, E sin cf>, in 
quadrature with the current. According to the proof given 
above, the average power produced by the quadrature component 
of the voltage is zero, so that the total average power is 

P ave = EI- cos <j>. r~. .... (77) 

A more rigid proof of this expression is given in problem 3 below. 

The product EI is called the apparent power, and cos <f> is 
referred to as the power-factor. Thus, the power-factor can be 
defined either as the cosine of the angle of phase displacement 
between the current and the voltage, or as the ratio of the true 
power P^e to the apparent power IE. The second definition 
is more general, because it applies also to non-sinusoidal currents 
and voltages. 

Referring to Fig. 13, the factor I cos <£ which enters into 
eq. (77) represents the projection of I upon the direction of the 
voltage OB, or E. Hence, eq. (77) can be interpreted by saying 
that the true power is equal to the product of the voltage by 
the component of the current in phase with it. This component 
of the current, I cos 4>, is therefore called the energy component, 
while the component I sin 0, at right angles or in quadrature with 
the voltage, is called the reactive component. 1 

Instead of resolving the vector of the current into two com- 
ponents, it is sometimes preferable to resolve the voltage E into 
1 The older name for this reactive component is wattless current. 



Chap. V] POWER IN ALTERNATING-CURRENT CIRCUITS 57 

the components E cos 4> and E sin 0, in phase and in quadrature 
with the current. In this case, eq. (77) is expressed in words by 
saying that the average power is equal to the current times the 
component of the voltage in phase with it. These components 
of the voltage are also called the energy component and the reactive 
component respectively. The two components of power, the true 
power EI cos (/>, and the reactive power EI sin 4>, stand in the 
same relation to the apparent power EI as the two sides of a 
right triangle bear to the hypotenuse; that is, 

(EIY = (EI cos 4>Y + (EI sin <f>y (78) 

Let now the current and voltage curves be different from pure 
sine-waves, and also different from each other in form. The 
fundamental equation 

Pave= (l/T) Cei-dt (79) 

holds true in all cases, so that if the curves are given graphically, 
the energy per cycle is found by multiplying the corresponding 
instantaneous values of e and i, and using the planimeter on the 
resultant curve. The average ordinate of this curve gives the 
average power. Of course, the parts of the resultant curve below 
the axis of abscissa? must be evaluated separately from those above 
it, and the difference of the two taken to represent the total 
energy. 

If the two waves are given in the form of Fourier series, an 
expression for the average power may be obtained in terms of 
the effective values of the harmonics. Substituting the expan- 
sions for e and i into eq. (79), two kinds of terms are obtained, — 
those containing products of two harmonics of the same frequency, 
and those containing products of two harmonics of different fre- 
quencies. The terms of the first kind, after integration, give re- 
sults of the same form as for the fundamental wave; that is, for 
the nth harmonic J E n I n cos <t> n , where E n and I n are the amplitudes 
of the nth harmonics, and <t> n is the phase displacement between 
them. The terms of the second kind give zero after integration, 
the proof of this being analogous to that in problem 3 of the pre- 
ceding article. Thus 

P ave = \ Exh cos 0i + i #3/3 cos <£ 3 + etc. . . (80) 

In other words, each harmonic contributes its own share of power, 
as if it were acting alone. 



58 THE ELECTRIC CIRCUIT [Art. 19 

Let E and i" be the effective values of some non-sinusoidal 
periodic voltage and current, measured, for instance, by means 
of hot-wire or dynamometer-type instruments. Let P ave be the 
average power according to eq. (80), or measured by a dyna- 
mometer-type wattmeter. Then the ratio P ave /EI is called the 
power-factor of the system, the same as with sinusoidal curves. 
This ratio is also often denoted by cos 0, meaning by the phase 
angle between the equivalent sine-waves of voltage and current, 
as defined in the preceding article. With the use of this angle and 
of the equivalent sine-waves, vector diagrams may be constructed 
and the corresponding calculations performed with currents and 
voltages deviating considerably from pure sine-waves, though of 
course such calculations check only approximately with the actual 
measurements. 

Prob. 1. Assuming the line current in problem 4, Art. 14, to be 452 
effective amperes, calculate the average power delivered by the alternator, 
and the power received at the opposite end of the line. 

Ans. 2444 kw.; 2350 kw. 

Prob. 2. Referring to problem 1, Art. 17, a wattmeter was connected 
into the heater circuit, and the true power was found to be 598 watts. 
Assuming all the three instruments to be in calibration, calculate the 
power-factor and the angle of displacement between the current and the 
voltage in the heater; also the energy component and the reactive com- 
ponent of the current. 

Ans. 95.4 per cent; 17° 30'; 5.39 amp.; 1.71 amp. 

Prob. 3. Deduce expression (77) for power by direct integration. 
Solution: Let the current be expressed by I m sin u; also let the voltage 
be leading by an angle </>, and therefore expressed as E m sin (u + <£)• 
Substituting these values into eq. (79), we get 

Pave = (ErJm/T) J sin u sin (u + 4>) dt, 

Jr»27r 
sin u [sin u cos <j> + cos u sin <t>] du, 
o 

Jf»2ir /»2T 

sin 2 u du + sin I sin u cos udu]. 
o Jo 

From a table of integrals we find that the value of the first integral is w, 
and that of the second is zero. Substituting these values, and introduc- 
ing the effective values of voltage and current, formula (77) is obtained. 
Prob. 4. Plot a sine-wave representing an alternating voltage of 
500 effective volts, and a current of the same frequency, of 20 effective 
amperes, lagging behind the voltage by 30 degrees. Plot on the same 
curve sheet the sine-wave of the instantaneous power, and check the 



Chap. V] POWER IN ALTERNATING-CURRENT CIRCUITS 59 

average ordinate of this curve with the value obtained by formula (77). 
Explain why the power is negative during a part of the cycle, remember- 
ing that there are local electromotive forces in the circuit. 

Prob. 5. Prove that the curve of power consists of a sine-wave of 
double frequency, plus a constant term, the latter representing the aver- 
age power. Compare with problem 3, Art. 16. Suggestion: ie = I m E m 
sin u sin (u + </>) . Use the trigonometric transformation, 2 sin A sin B = 
cos (A - B) - cos (A + B). 

Prob. 6. A non-sinusoidal voltage is represented by the equation 
e = 270 sin u + 62 sin 3 (u + 15°) + 16 sin 5 {u — 25°) ; the correspond- 
ing line current is i = 18 sin (u — 30°) — 7 sin 3 (u + 50°) + 2.5 sin 
5(u + 10°). Calculate the true average power and the power-factor of 
the system. 

Ans. P ave = \ (4860 cos 30° - 454 cos 75° - 40 cos 5°) = 2027 watts; 
the power-factor is 75 per cent. 



CHAPTER VI 

INDUCTANCE* REACTANCE AND IMPEDANCE 

20. Inductance as Electromagnetic Inertia. Experiment 
shows that an electric current in a variable state behaves as if 
it possessed inertia; there is an opposition to any change in its 
magnitude and direction. This opposition is manifested in the 
form of an " induced" electromotive force in such a direction as 
to tend to counteract the change in current. Thus, if an external 
e.m.f. tends to increase the current, the induced e.m.f. is in a 
direction opposite to that of the current; but when, on the other 
hand, the current for some reason decreases, the induced e.m.f. 
is in the same direction as the current, and therefore tends to 
strengthen it. These reactions of the current are similar to those 
exerted by a moving body; for instance, the water in a pipe, when 
its motion is accelerated or retarded. In practical applications, it 
is convenient to consider, not the reactions themselves, but the 
external forces necessary to overcome them. Thus, in the case 
of a moving body of mass m, the external force necessary to com- 
municate to it an acceleration dv/dt is F = m dv/dt. Here F 
is positive when the acceleration is positive, and vice versa. 
Similarly, to increase a current at a rate of di/dt, an external e.m.f. 
is necessary of the magnitude 

e = Ldi/dt, (81) 

where L is a constant which characterizes the circuit and is analo- 
gous to the mass m in the mechanical motion. The coefficient L 
is called the inductance of the circuit, and depends upon its shape 
and proportions, the presence or absence of iron, the number of 
turns which the conductor makes, and some other factors, which 
it is not necessary to discuss here. In most books the right-hand 
side of eq. (81) is written with the sign minus, because e is under- 
stood to mean the induced e.m.f. or the reaction of the circuit; 
while in our case e designates the external voltage, equal and 

60 



Chap. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 61 

opposite to this reaction. The form of the equation used here is 
preferable, because in practice one deals with components of the 
applied voltage rather than with the induced counter-e.m.f.; 
moreover, the minus sign is apt to confuse a beginner. 

The inertia effect of the electric current is brought about 
through the mechanism of the magnetic field produced thereby. 
When the current varies, the flux embraced by the electric circuit 
also changes, and according to Faraday's law of induction this 
flux induces in the circuit an e.m.f. Thus, postulating the exist- 
ence of electromagnetic inertia, and stating the law of induced 
e.m.f., are perhaps but two different ways of expressing the same 
physical phenomenon, the true nature of which is at present un- 
known. Any arrangement of the circuit which increases the flux 
linked with it, also increases its inductance L or the inertia effect. 
The inductance of a given electric circuit can be calculated with 
more or less accuracy, 1 or it can be measured experimental^, 
using eq. (81). For our present purposes we shall assume L to 
be a constant quantity, which characterizes the inertia of a given 
electric circuit, according to eq. (81), without any reference to 
the nature of the magnetic flux which produces it. Mechanical 
inertia is used in physics and in engineering as a fundamental 
entity, without explaining it in any other terms, while the mystery 
as to its cause is just as deep as that surrounding the electro- 
magnetic inertia. Some modern physicists even believe that all 
inertia is of an electromagnetic nature. 

The fact that a body resists acceleration, together with the 
law of conservation of energy, leads to the conclusion that a mov- 
ing body possesses a certain amount of stored energy. The 
external work done upon a body while it moves through a dis- 
tance ds is F • ds = m(dv/dt)ds, or, since ds = v • dt, we have 
F • ds = mv ' dv. The total work done upon the body while accel- 
erating it from rest to a velocity v is therefore 



W 



Jf*s r»v 

F ds = I mv • dv = J mv 11 
o Jo 



According to the law of conservation of energy, this work is 
stored in the moving body as its kinetic energy. 

The electrical work done in increasing a current against the 
induced electromotive force, during the time dt is dW = ei • dt, or 
1 See the author's Magnetic Circuit, Chapters 10 to 12, 



62 THE ELECTRIC CIRCUIT [Art. 20 

substituting for e its value from eq. (81), dW = Lidi. The total 
energy supplied to the circuit from the external source of power, 
while the current increases from zero to a certain value i, is 



W 



= C l Lidi = \Li 2 (82) 



This does not include the energy required for^supplying the i 2 r 
loss. According to the law of conservation of energy, expression 
(82) represents the energy stored in the circuit as long as the value 
of the current remains the same. When the circuit is broken, 
this energy is converted into heat. Analogously, when a non- 
elastic moving body is stopped, its accumulated energy is con- 
verted into the heat of impact. Inductance can be defined from 
either eq. (81) or (82); and for most purposes the two definitions 
are identical. Similarly, in mechanics, mass may be defined 
either as the ratio of F to dv/dt, or as a ratio of the kinetic energy 
toi^ 2 . 

The unit of inductance in the ampere-ohm system is called 
the henry. According to eq. (81), a circuit has an inductance of 
one henry when one volt is necessary in order to increase the 
current at a rate of one ampere per second. -This one volt does 
not include, of course, the e.m.f. necessary for overcoming the 
resistance of the circuit. The henry being rather a large unit, 
inductance is frequently measured in millihenrys. Substituting 
into eq. (81) the physical dimensions of the voltage in the ampere- 
ohm system, we get [IR] = [LI/T] or [L] = [RT]. In other 
words, the henry stands for the " ohm-second." For this reason, 
one instrument for measuring inductance has been called by its 
inventors "the secohmmeter. " 

All actual circuits which possess inductance, at the same time, 
have some resistance, however small it may be. Therefore, the 
total instantaneous voltage applied during a variable state is 

e = ir + L di/dt (83) 

Ohmic resistance may be compared to mechanical friction, 
so that eq. (83) can be interpreted by reference to the mechan- 
ical analogy used above, in the following way; namely, the 
force necessary to accelerate a body must be augmented in 
practice by the amount required for overcoming the inevitable 
friction. 



Chap. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 63 

Prob. 1. A circuit which possesses an inductance of 12 millihenrys 
and carries a direct current of 150 amp. is broken within one-fifth of a 
second. What is the average voltage induced in the circuit during this 
interval of time? Ans. 9 volts. 

Prob. 2. Calculate the electromagnetic energy stored in the circuit of 
the preceding problem while the current is steady. Ans. 135 joules. 

Prob. 3. The current in a coil is made to vary at a uniform rate of 
250 amp. per second. At the instant when the current is equal to 150 
amp., a voltmeter connected across the terminals of the coil reads 295 
volts; when the instantaneous current is 100 amp. the voltmeter reading 
is 230 volts. From these data calculate the resistance and the induc- 
tance of the coil. Ans. 1.3 ohms; 0.4 henry. 

21. Reactance. It is natural to expect the inductance to 
exert a considerable influence upon the voltage and current rela- 
tions in an alternating-current circuit, because the current is 
varying in magnitude all the time. The influence of inductance 
in this case is analogous to that of the inertia of the moving parts 
in a reciprocating engine; i.e., energy is stored during the periods 
of increase in velocity (or in current) , and is returned to the source 
of power during the intervals of time when the velocity (or the 
current) decreases. There is no net gain or loss of energy for a 
complete cycle, although the instantaneous values of current and 
voltage may be considerably affected. 

Consider first a part of a circuit which has inductance only, 
the resistance being negligible. Let the current vary according 
to the familiar law i = I m sin (2 wft — a) . Substituting this value 
into eq. (81), we get 

e = 2 it f LIm cos (2 irft - a)., . . . . (84) 

which means that the voltage necessary to force a sinusoidal 
current through an inductance also varies according to the sine 
law, and is in leading quadrature with the current. The ampli- 
tude of the voltage E m = 2 irfLI m} or the relation between the 
effective values of voltage and current, is 

E = 2irfLI (85) 

It will be seen from this relation that, in alternating-current cal- 
culations, the quantities / and L always appear as a product. 
It is therefore convenient to introduce, for the sake of abbrevia- 
tion, a new composite quantity x, defined by the relation 

x = 2wfL (86) 



64 



THE ELECTRIC CIRCUIT 



[Art. 21 



The quantity x is called the reactance of the circuit, and always 
refers to a stated frequency /. Equation (85) becomes then 

E = xl, (87) 

from which it follows that reactance is measured in ohms, like 
resistance. This does not mean, however, that the two quanti- 
ties are similar in their physical nature. 

r Let now some resistance be con- 

nected in series with the inductance 
(Fig. 18), or let the coil which pos- 
sesses inductance have also an appre- 
ciable resistance. Substituting the 



MAMAAAAAn 



X=2irfL 



Fig. 18. Resistance and re- expression for i, given above, into 



actance in series. 



eq. (83), we get 



e = rl m sin (2 irft — a) + xl m cos (2 irft — a). 



(88) 



Since the sum of two sine- waves is also a sine- wave (see Art. 14), 
the total voltage e varies according to the sine-law. These com- 
ponent sine-waves of voltage, their sum, and the current wave 
are shown in Fig. 19. The student is advised to study this figure 




Fig. 19. The instantaneous current and voltage relations in the circuit 
shown in Fig. 18. 



very carefully, because it represents one of the most important 
fundamental relations in the whole theory of alternating currents. 
The same relations are represented vectorially in Fig. 20, and the 
two figures may be conveniently examined together. One de- 
scribes the phenomenon from instant to instant; the other gives 



Chap. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 65 

the salient features in a symbolic form. The vector E consists 
of one component It in phase with the current, and another Ix 
in leading quadrature with the current. The first component 




Fig. 20. 



I?' T 

The current and voltage relations in the circuit shown in Fig. 18, 
represented vectorially. 



serves to overcome the ohmic resistance; the second, the reac- 
tance of the circuit. From the triangle of voltages we have 
E 2 = p r 2 _j_ p x ^ 

or E=lVr* + x* (89) 

For the phase displacement between the current and the voltage 
we have 

tan 4> = Ix/Ir = x/r, (90) 

or the power-factor 

cos 4> = r/Vr 2 + x 2 (91) 

The hydraulic analogue shown in Fig. 21 may make these 
relations clearer. ACDGA represents a closed pipe circuit in 
which water is made to oscillate 
to and fro by means of the piston 
B. The water is assumed to be 
devoid of inertia, and the inertia 
of the whole circuit is concen- 
trated in a heavy mass F, which 
moves freely with the water. 
The force upon the piston rod 
H is analogous to the alternat- 
ing voltage E in Fig. 18; the 
velocity of the water is analo- 
gous to the alternating current, the friction in the pipes represents 
the ohmic resistance r, and the inertia of the heavy mass F stands 
for the inductance L. To make the analogy closer, we assume 
that the piston is forced to perform a simple harmonic motion; 




Fig. 21. 



A hydraulic analogue 
to Fig. 18. 



66 THE ELECTRIC CIRCUIT [Art. 21 

so that the velocity of the water varies with the time according 
to the sine law, and may be represented by the curve for i in 
Fig. 19. 

The force upon the piston B consists of two parts, that re- 
quired for overcoming the friction in the pipes, and that necessary 
for accelerating and retarding the mass F. These two compo- 
nents of the force can be represented by the curves for ir and ix 
in Fig. 19. The frictional reaction is at a maximum when the 
piston is in the middle of its stroke, because there the velocity 
of the water is the greatest. On the other hand, the acceleration 
is zero in this position, so that the mass F exerts no reaction. 
At the ends of the stroke the acceleration or retardation is at a 
maximum, so that the force necessary for constraining the mass 
F to the prescribed motion is at a maximum; however, the fric- 
tional resistance is equal to zero. Adding the two sinusoidal com- 
ponents, we find the resultant force upon B, corresponding to 
the curve e in Fig. 19. It will be seen that e reaches a maximum 
before the center of the stroke; this gives a phase angle between 
the force and the velocity that is analogous to the phase angle 
between the voltage and the current. The student can easily 
deduce that the force leads the velocity in phase, and that the 
displacement is greater the larger the mass F, as compared to 
the frictional resistance; in other words, the greater the reactance 
as compared to the resistance. It may be shown also that the 
inertia reaction of the same mass F is greater for a higher fre- 
quency of oscillation, because the acceleration and retardation are 
proportionately larger. 

Prob. 1. The inductance of a coil is 0.2 henry; its ohmic resistance 
is negligible. Draw a curve giving the voltage necessary to maintain a 
current of 12 amp. through the coil, at frequencies ranging from zero to 
100 cycles per second. 

Ans. A straight line through the origin; at / = 100, E = 1508 volts. 

Prob. 2. A reactive coil without iron draws a current of 75 amp. when 
connected across a 110- volt 25-cycle circuit. What current would it draw 
at 60 cycles and at the same voltage, provided that the effect of its 
resistance can be neglected? Plot a curve of current at intermediate 
frequencies. 

Ans. 31.25 amp.; equilateral hyperbola asymptotic to both axes. 

Prob. 3. The reactive magnetizing current of a 2200-volt, 600-kilo- 
watt, 50-cycle transformer must be not over 2.5 per cent of the full-load 
current. What is the lower limit of its no-load reactance and inductance? 

Ans. 322.5 ohms; 1.027 henrys. 



Chap. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 67 

Prob. 4. The coil considered in problem 1 is connected in series with 
a 100-ohm resistance; it is required to maintain a current of 12 amp. 
through the two, at various frequencies. Supplement the curve obtained 
in that problem with curves of voltage drop across the resistance, and the 
total voltage across the combination. Plot also the corresponding values 
of power-factor. Determine the ordinates of the curves graphically, 
and check a few points analytically. 

Ans. E r = 1200 volts, independent of the frequency. At / = 0, 
E totm =Er, and cos 0=1. At /= 100, E total = 1927 volts, cos <f> = 62.25 
per cent. 

Prob. 5. Three simultaneous instrument readings in a power house 
are: 7520 kw.; G6kv.; 147 amp. The power-factor meter shows that 
the current is lagging behind the voltage. What, are the readings at the 
same instant at the receiving end of the line, if its resistance is 45 ohms 
and its reactance 83 ohms. Hint: Draw the vectors of the generator 
voltage and current in their true relative position. Subtract the ohmic 
drop in phase with the current, and the reactive drop in quadrature with 
it. The result will give the receiver voltage in its true magnitude and 
phase position. Ans. 6547 kw.; 53.4 kv. 

Prob. 6. In order to determine the power input into a single-phase 
110-volt motor, without the use of a wattmeter, the motor is connected 
in series with a non-inductive resistance across a 220-volt circuit. The 
resistance is adjusted so that the voltage across the motor terminals is 
110, when the motor is carrying the required load. Under these condi- 
tions the voltage across the resistance is found to be 127, and the current 
through the motor 23 amp. From these data determine graphically the 
power-factor of the motor, and calculate its power input. 

Ans. 72.3 per cent; 1826 watts. 

Prob. 7. Referring to the preceding problem, calculate cos <j> trigo- 
nometrically, from the triangle of voltages, instead of determining it 
graphically. 

22. Impedance. When a reactance is connected in series 
with a resistance, eqs. (89) and (91) indicate that the current and 
voltage relations are determined, not by the value of the reac- 
tance alone, but by a composite expression 

2 = Vr 2 + x\ . (92) 

The quantity z has the dimension of a resistance, and is called 
the impedance of the circuit. It can hardly be called a physical 
quantity, but rather an abbreviation for a certain combination 
of the physical properties of a circuit; in other words, an abbre- 
viation for the radical in eq. (92). Introducing the value of z 
into eqs. (89) and (91), we obtain 

E = zl (93) 

and cos = r/z (94) 



68 



THE ELECTRIC CIRCUIT 



[Art. 22 



Impedance may be denned from eq. (93) as the ratio of the voltage 
to the current in a circuit containing resistance and reactance. 
In a non-inductive circuit the impedance is simply equal to the 
total resistance, while in a purely inductive one the impedance 
is equal to the reactance. It must be clearly understood that 
eq. (93) gives only the relation between the magnitudes of the 
vectors. The phase relation is given by Fig. 20, or by eq. (94). 
The three quantities r, x, and z form a triangle of which z is 
the hypothenuse (Fig. 20). This triangle is similar to the tri- 
angle of voltages, but the quantities r, x, and z are not vectors in 
the same sense as currents and voltages are. From the impe- 
dance triangle we have the following useful relations: 

r = z cos ^ (95) 

x = z sin (96) 



and 



i < E, > 



When two impedances are connected in series (Fig. 22), the 
voltage and current relations are as represented in Fig. 23. The 

total terminal voltage E is less 
than the arithmetical sum of 
the voltages Ex and E 2 across 
the two impedances, and is 
equal to their geometric sum. 
The resultant phase angle <£ has 
a value intermediate between 
the phase angles <£i and <p 2 of the two component impedances. 
It will be seen from the triangle ABC that the resultant voltage 



Fig. 22. Two impedances in series. 




£5-1 



Fig. 23. The current and voltage relations in the circuit shown in Fig. 22. 



is the same as that required by an impedance which consists 
of a resistance r x + r 2 and a reactance x\ + x 2 . In other words, 



Chap. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 69 

the resultant impedance is 

z = V(n + r 2 ) 2 + Oi + xtf, .... (97) 
and the resultant phase angle is determined from the equation 

tan 4>=(xi + x 2 )/(r 1 + r 2 ) (98) 

These equations show that two impedances are added in series 
by adding the resistances and the reactances separately. An 
impedance of 5 ohms in series with one of 7 ohms is not equal to 
an impedance of 12 ohms, but as a rule is less. The relations 
shown in Figs. 22 and 23, and eqs. (97) and (98), are easily ex- 
tended to any number of impedances in series. Dividing all the 
voltage vectors in Fig. 23 by the value of the current I, the dia- 
gram of voltages is converted into one of impedances, as in Fig. 20, 
the relations being represented by eqs. (97) and (98). It must be 
borne in mind, however, that from a physical point of view the 
latter relations are not vectorial in the same sense as are those 
of the voltages. 

Prob. 1. The impedance of a coil is 7.5 ohms at 60 cycles; the re- 
sistance measured with direct current is 6 ohms. What is the inductance? 

Ans. 11.9 millihenrys. 

Prob. 2. Two impedance coils are connected in series across a 292- 
volt line. The voltages across the coils are 152 and 175 respectively; 
the current is 7.3 amp. Knowing that the resistance of the first coil is 
10 ohms, determine graphically the resistance of the second; also the 
impedances of both coils. 

Ans. r 2 = 12.95; Zx = 20.82; z 2 = 23.97, all in ohms. 

Prob. 3. When a certain non-inductive resistance is connected across 
a source of alternating voltage, a current / flows through it. When an 
inductance, containing negligible resistance, is connected across the same . 
source of voltage, the current is V. What are the current and the phase 
displacement when the resistance and the inductance are connected in 
series across the same source? Solution: Let the unknown voltage be 
E. The unknown resistance is r = E/I; the unknown reactance x = 
E/T. When the two are connected in series, the impedance z = [(E/I) 2 
+ (E/V) 2 ]*. Consequently, the current is E/.z = IV /(I 2 + I' 2 ) * ; tan </» = 
x/r = I IV. 

23. Influence of Inductance with Non-sinusoidal Voltage. 

(a) Let an alternating voltage e of an irregular form, such as is 
shown in Fig. 14, be applied at the terminals of a pure resistance r 
(non-inductive). The current through the resistance is at any 
instant equal to e/r, and consequently has the same wave form 
as the voltage. 



70 THE ELECTRIC CIRCUIT [Art. 23 

(b) Let now the same voltage be applied at the terminals of 
a pure inductance L (without resistance). It may be said a priori 
that the current wave will be different from that of the voltage, 
and will approach more nearly a sine-wave. This follows from the 
very concept of inductance as the inertia of the circuit; the high- 
frequency harmonics in the voltage are unable to produce currents 
of the same magnitude as at lower frequencies, because the react- 
ance offered to each harmonic is proportional to its frequency. 
This property of an inductance of choking higher harmonics is 
useful in some applications. 

Let the voltage across an inductance be given in the form of a 
Fourier series, 

e = E l sin (2 irft - on) + E 3 sin 3 (2 irft - a 3 ) + etc. 

Substituting its value in the fundamental eq. (81), we get 

#i sin (2 wft - ai) + E 3 sin 3 (2 irft - a 3 ) + etc. = L di/dt. 

Multiplying both sides of this equation by dt and integrating 
gives 

- (#i/2 irf) COS (2 irft - ai) - (#3/6 irf) COS 3 (2 irft - a z ) 
— etc. = Li + const. 

The constant of integration is equal to zero, because the current 
cannot have a unidirectional component without a commutating 
device or electric valve of some sort. Therefore 

i = - (#1/2 tt/L) cos (2 irft - ai) - (#3/6 tt/L) cos 3 (2 irft - a 3 ) 

- etc., (99) 

which means that each harmonic in the e.m.f. produces its own 
current, as if this harmonic were acting alone. The total current 
is the sum of such harmonic currents. The reactance of the coil 
for the nth harmonic is n times as great as for the fundamental 
wave; therefore, the higher harmonics in the current are rela- 
tively smaller than those in the voltage wave. 

(c) Let now a non-sinusoidal alternating voltage be impressed 
at the terminals of an impedance coil, and let it be required to 
determine the wave form of the current. The result to be 
expected will be intermediate between those derived for a pure 
resistance and a pure inductance; viz., the current wave will be 
more nearly of sine form than the voltage wave, but not to the 
same extent as in the case of a pure inductance. 

Substituting the above given expansion for the voltage wave 



Chap. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 71 

into the fundamental eq. (83), we obtain a differential equation for 
i, which equation some readers may not be able to solve. We 
choose, therefore, the opposite way; that is, we assume the current 
wave to be given, instead of the voltage wave, and determine the 
corresponding voltage wave from eq. (83). This procedure is 
much simpler, because it involves differentiation instead of inte- 
gration. Let the current be given in the form 

i = 1 1 sin (u — «i) + 1 3 sin 3 (u — a 3 ) + etc., where u = 2 irft. 

Substituting this value into eq. (83) and rearranging the terms, 
gives 

e = [IiV sin (u — ai) + 2 t/LIi cos (u — a{)] +[!& sin 3 (u — a 3 ) 

+ Gt/LIz cos 3 (u - a 3 )] + etc. . . . (100) 

This result shows that each harmonic of the current requires a 
corresponding harmonic of the voltage, as if it were flowing alone. 
The total voltage is equal to the sum of the harmonic voltages. 
Therefore we conclude that, conversely, if the voltage were given, 
the current would be equal to the sum of the harmonic currents 
produced by the respective harmonics in the voltage. If the 
impedance to the first harmonic is Z\ = Vr 2 + x 2 , that to the 
third harmonic is z 3 = vV + (3 x) 2 , and in general the impedance 
to the nth harmonic is z n = Vr 2 + (nx) 2 . The phase displace- 
ment between the corresponding harmonics of current and voltage 
is determined from the condition, tan 4> n = nx/r, or cos <f> n = r/z n . 
The general conclusion reached is as follows : When the applied 
voltage contains higher harmonics, the total current is found by 
summing the harmonic currents due to each harmonic of the 
voltage acting alone. 

Prob. 1. The effective value of the fundamental wave of an e.m.f. is 
110 volts; it has a pronounced third harmonic, of 24 per cent of the fun- 
damental wave. This voltage is applied across a pure reactance, equal to 
5 ohms for the fundamental frequency. Calculate the current. 

Ans. 22.07 amp. 

Prob. 2. An alternating voltage is represented by the expression 
170 sin 250 t + 62 sin (1250 t + 2.3). It is applied to an impedance coil 
having an inductance of 45 millihenrys and a resistance of 7 ohms. Show 
that the current in amperes is equal to 12.82 sin (250 t — 1.015) + 
1.09 sin (1250 £ + 0.853). 

24. The Extra or Transient Current in Opening and Clos- 
ing a Circuit. Since an electric current possesses inertia in 



72 THE ELECTRIC CIRCUIT [Art. 24 

the form of inductance, no current can be established or broken 
instantly, unless the applied electromotive force be infinitely 
large. Thus, when a large electromagnet is connected to a source 
of continuous voltage, the current increases during an appreciable 
interval before.it reaches its final value. Again, when the circuit 
is broken, the current continues in the form of an arc through 
the air for an appreciable time. In a majority of cases these 
transient phenomena at the opening and closing of a circuit are of 
no practical importance, yet there are circumstances under which 
they must be taken into consideration; for instance, in switching 
on and off large amounts of energy, in high-frequency oscillations, 
in highly inductive circuits, etc. We shall consider here two 
simple cases of such extra currents; namely, when a circuit pos- 
sessing resistance and inductance is connected to a source of 
(a) continuous voltage and (b) sinusoidal alternating voltage. 

(a) Direct Voltage. When e in eq. (83) is constant, one value 
of i which satisfies this equation is i = e/r, because in this case 
di/dt = 0. However, this is not the most general solution, 
because it is possible to select an exponential expression in addi- 
tion to the constant i, which will satisfy the equation. Put 

i = e/r + Ce~^, (101) 

where e is the base of natural logarithms, and C and r are certain 
constants. Substituting this value of i into eq. (83), we get 

Cre~^ - (L/r)Ce"^ = 0, 
or 

r = L/r. 

Besides, i = when t = 0, so that expression (101) becomes 
= e/r + C, from which 

C = -e/r, 
and consequently 

i = (e/r) (1 - cr tr/L ) (102) 

In other words, when a direct-current circuit is closed, the current 
increases at first rapidly, then more and more slowly; and theo- 
retically it reaches its final value of e/r only after an infinite time. 
In reality, the current becomes practically constant after a frac- 
tion of a second, unless the inductance is exceedingly large. The 
factor r = L/r is called the time constant of the circuit; it deter- 
mines the rate of the initial rise in current, and has the dimension 
of time. 



Chap. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 73 

(b) Sinusoidal Voltage. If the voltage follows the law e = 
E m sin2irft, one solution of eq. (83), as we have seen before, is 
i = (E m /z) sin (2 wft — 0), where cos = r/z. But this is not the 
most general solution, because it is possible to add to it an expo- 
nential term of the form Ce~' /T , and to select the time constant r 
in such a way that this term will cancel in eq. (83). Since the 
sine term of the current alone satisfies the equation, we will find 
as before r = L/r. The constant C is determined by the condi- 
tion that i = when t = 0, or 

0= -(EJz)smct> + C, 
from which 

C = (E m /z) sin 4> = E m x/z\ 

Therefore the current 

i = (EJz) sin (2irft - <f>) + (E m x/z')e-^ L . . . (103) 

Under ordinary conditions the exponential term becomes negli- 
gibly small within a fraction of a second, so that it is legitimate 
to consider the current to be a pure sine-wave, as we have done 
heretofore. However, the extra current may be of importance in 
transient phenomena, for instance, at the moment of closing a 
circuit. 

The solutions (102) and (103) of eq. (83) are found above by 
trials, because it is assumed that the reader is not familiar with 
the general method for the solution of linear differential equations ; 
otherwise, the solution could have been written directly. Equa- 
tion (83) is of the form 

dij/dx + Py = Q, (104) 

where P and Q are functions of x or constants. By referring to 
any book on differential equations, the reader will find that the 
general solution of this equation is 

y = e-°{Je"Qdx + 6\, .... (105) 
where 



/ 



Pdx (106) 



Prob. 1. The current in a coil due to a constant e.m.f. reaches 99 per 
cent of its final value within one hundredth of a second after the circuit 
is closed. Show that the time constant of the coil is equal to 2.17 milli- 
seconds. 



74 THE ELECTRIC CIRCUIT [Art. 24 

Prob. 2. Show that the time constant may be denned as the interval 
of time during which the current reaches (e — l)/e = 0.632 of its final 
value. 

Prob. 3. Select the constants of an alternating-current circuit so as 
to have a power-factor of about 80 per cent; and plot curves of (a) the 
voltage, (b) the sinusoidal component of the current, (c) the exponential 
component of the current, and (d) the total current, for the first few 
cycles after the circuit is closed. 

Prob. 4. Extend the theory given above to the case where the cir- 
cuit is closed at an instant when the alternating voltage is not equal to 
zero. 

Prob. 5. Check the solutions (102) and (103), using formula (105). 

Prob. 6. When an impedance, consisting of r and L, is suddenly short- 
circuited, so that e becomes instantly equal to zero, show that the line 
current gradually disappears according to the exponential law i = i {) e~ rt/L , 
where u is the magnitude of the current at the instant of short-circuit. 



CHAPTER VII 

SUSCEPTANCE AND ADMITTANCE 

25. Concept of Susceptance. The concept of reactance, as 
introduced in Art. 21, indicates the degree of difficulty in forcing 
an alternating current through a coil, against the reaction of an 
alternating magnetic field. In this respect, reactance is analo- 
gous to resistance. We have seen, however, in Chapter I, that it 
is more convenient to use conductances, when resistors are con- 
nected in parallel. Similarly, when reactive coils or reactors are 
connected in parallel, it is more convenient in calculations to use 
the reciprocals of their reactances. The reciprocal of reactance is 
called susceptance, and is usually denoted by the symbol b. Thus, 
by definition, the susceptance 

b= l/ x = 1/(2 tt/L) (107) 

By analogy with conductance, one may say that the susceptance 
measures the degree of ease in forcing an alternating current 
through a coil, against the reaction of a pulsating magnetic field. 
Since reactance is measured in ohms, susceptance is measured in 
mhos. Equation (87) becomes 

I = bE, (108) 

it being understood as before that the current lags by 90 degrees 
behind the voltage. The student is reminded that the concept 
of susceptance, like that of reactance, implies pure inertia reaction, 
without any ohmic resistance; this limitation is very important 
for a clear understanding of the rest of the chapter. 

When several inductive coils are connected in parallel, their 
susceptances are simply added together, or 

b eq = 6i + b 2 + etc (109) 

The proof is similar to that for the addition of conductances (see 
Art. 3). Thus, a susceptance of 3 mhos in parallel with one of 
2 mhos gives a total susceptance of 5 mhos. 

75 



76 



THE ELECTRIC CIRCUIT 



[Art. 26 



Prob. Two reactive coils of 10 and 20 millihenrys respectively are 
connected, first in series and then in parallel, across a 40-cycle, 180-volt 
line. The ohmic resistance of the coils is negligible. What is the current 
in each case? Ans. 23.85 amp.; 107.35 amp. 

26. Concept of Admittance. Let now a pure inductance be 
connected in parallel with a pure ohmic resistance, across a source 
of alternating voltage E (Fig. 24), and let it be required to find 




Fig. 24. A susceptance in parallel 
with a conductance. 



A ^^ 




5» 
HI 

II 






Fig. 25. The voltage and current 
relations in the circuit shown 
in Fig. 24. 



the total current through the combination. The inductance can 
be expressed as a susceptance, and the resistance as a conductance. 
The current through the susceptance, according to eq. (108), is 
bE, in quadrature with the voltage (Fig. 25) ; the current through 
the conductance, according to eq. (2), is gE, in phase with the 
voltage. The total current 

/ = V(Eg) 2 + (Eb) 2 = E Vg 2 + b 2 , . , . (110) 
and the phase angle is determined from the relation 

ttmcf> = Eb/Eg = b/g (Ill) 

or 



cos0 = g/Vg 2 + b 2 . . . . . . (112) 

In the case of a series connection, we have found it convenient 
to introduce the impedance 2 as a symbol for Vr 2 -\- x 2 . Simi- 
larly, in a parallel connection it is convenient to introduce the 
abbreviation 

y = Vtf~+b 2 . ...... (113) 

The quantity y is called the admittance of a circuit, and is measured 
in mhos, the same as b and g. Equation (110) becomes 

I = yE, (114) 

and eq. (112), 

cos0 = g/y. . (115) 



Chap. VII] SUSCEPTANCE AND ADMITTANCE 



77 



The three quantities g, b and y form a triangle (Fig. 25), in which 
y is the hypothenuse, and the angle adjacent to g is the phase 
angle </>. From this triangle we obtain two useful relations, 

and V = y™*) (n6) 

o = y sin </>. J 

When there are several susceptances and conductances in parallel 
(Fig. 26), the reactive and the energy components of the current 




Fig. 26. Susceptances and conductances in parallel. 

must be added separately (Fig. 27). Therefore, the amperes per 
volt in phase or the conductances, and the amperes per volt in 



E 


E&2 




E&3 




ng 3 










<h 


Eflf, 





Fig. 27. The voltage and current relations in the circuit shown in Fig. 26. 

quadrature or the susceptances, must also be added separately, so 
that the equivalent admittance 

V = VCflf! + g 2 + etc.) 2 + (&! + b 2 +- etc.) 2 , . (117a) 
and 

tan = (&! + b 2 + etc.)/(gi + flf 2 + etc.). . . . (117b) 

The student should compare Fig. 27 with Fig. 23 in order to see 
the similarity of procedure and the difference in the physical 
phenomena in the two cases. With a series connection, it is the 
current that is common to all the parts of the circuit, while the 
partial voltages are added geometrically. In a parallel combina- 
tion, the voltage is common to all the branches, while the com- 
ponent currents are combined in their proper phase relations. 



78 



THE ELECTRIC CIRCUIT 



[Art. 27 



The following table gives the quantities denned in this and the 
preceding chapter, in their proper relations. 



Friction 
Effect 


Inertia 
Effect 


Connection 


Result 


Unit 


Resistance r 
Conductance g 


Reactance x 
Susceptance b 


Series 
Parallel 


Impedance z 
Admittance y 


Ohm 
Mho 



Prob. 1. What susceptance must be connected in parallel with a 
resistance of 0.2 ohm, in order to bring the power-factor of the combina- 
tion down to 80 per cent? Also, what is the value of the resultant ad- 
mittance? Ans. 3.75 mhos; 6.25 mhos. 

Prob. 2. Two electrical devices are connected in parallel to a line of 
voltage E. One device consumes a current 7i at a power-factor cos <fn; 
the total line current is I, lagging behind the voltage by an angle <f>. 
Show how to determine graphically the susceptance and the conductance 
of both devices. 



27. Equivalent Series and Parallel Combinations. Let a 

resistance r 8 be connected in series with a reactance x a ; also let 
another resistance r p be connected in parallel with a reactance x p . 
If the values of the resistances and reactances are so selected that 
the series combination, when connected to the same source of 
supply, will let through the same current at the same power- 
factor as the parallel combination, then the two combinations are 
called equivalent. It is sometimes convenient to replace a given 
series combination by an equivalent parallel combination, and 
vice versa. For instance, when some parts of a circuit are in 
parallel and others in series, it is convenient for numerical cal- 
culations to replace them all by an equivalent parallel or series 
combination. 

The problem is to find the relation between the four quanti- 
ties r s , r p , x s and x p , if these quantities form two equivalent com- 
binations. According to the above-given definition, the angle 
is the same for both, and besides, according to eqs. (93) and (114), 

y = l/z, ....... (118) 

where y refers to the parallel combination and z to the equiva- 
lent series combination. Combining now eqs. (116), (95) and (96), 
we have 

l/r p = g = ycos<f> = (1/z) (r s /z) = r s /z 2 ; 

l/xp = b = y sin <j> = (1/z) (x s /z) = x s /z 2 ; 



Chap. VII] SUSCEPTANCE AND ADMITTANCE 79 

or 

r 8 r p = z* = 1/V; (119) 

x 8 x p = z* = 1/y* (120) 

By means of eqs. (119) and (120) a series combination can be 
replaced by an equivalent parallel combination, and vice versa. 
Instead of r p and x p , their reciprocals, g and b, may be used. In 
practice, g and b are usually spoken of as the conductance and the 
susceptance of either the series or the parallel combination; but 
it must be clearly understood that they are the reciprocals of r p 
and x p , and not of r 8 and x 8 . If r a and x a are given, it is first 
necessary to determine r p and x p from eqs. (119) and (120), and 
then to take their reciprocals. In other words, for a series circuit 
the equivalent conductance and susceptance are 

Q = r 8 /z\ (121) 

and b = x 8 /z 2 (122) 

On the other hand, if g and b are given, 

r* = g/y 2 ; (123) 

x 8 = b/if (124) 

The reciprocals of r 8 and x 8 are of no practical importance, and 
are not used in this work. 

Prob. 1. An impedance coil has a reactance of 7.5 ohms; the resist- 
ance of the winding is 2 ohms. What are the susceptance and the con- 
ductance of the equivalent parallel combination? 

Ans. 124.3 and 33.2 millimhos. 

Prob. 2. Check the answer to the foregoing problem by actually cal- 
culating the current and the power-factor of the series and the parallel 
combinations at some assumed voltage. 

Prob. 3. Show that r p and x p are always larger than r 8 and x 8 re- 
spectively. Hint: In eqs. (119) and (120) replace z 2 by r 8 2 + x 8 2 , and 
solve for r p and x p . 

Prob. 4. An apparatus takes 25 amp. and 2000 watts at 110 volts, 
the current being a lagging one. What are the equivalent conductance 
and susceptance of the device? What are the resistance and reactance 
in series equivalent to this apparatus? 

Ans. 0.165 mho; 0.156 mho; 3.2 ohms; 3.04 ohms. 

Prob. 5. In adjusting a measuring instrument, a non-inductive re- 
sistance of 120 ohms was used in parallel with a choke coil. The imped- 
ance of the coil was 75 ohms, its resistance 16 ohms. In the regular 
manufacture of the instrument it is desired to use a resistance and a re- 
actance in series. Determine their values, either graphically or analyti- 
cally. Ans. r s = 37.9 ohms; x 8 = 44.15 ohms. 



80 THE ELECTRIC CIRCUIT [Art. 28 

28. Impedances in Parallel and Admittances in Series. In 

the preceding chapter we have learned how to add impedances 
in series, and in this chapter how to add admittances in parallel. 
Let now two or more impedances be connected in parallel, and let 
it be required to find the equivalent impedance. This is done 
by replacing each of the given impedances by an equivalent 
parallel combination, and then adding their admittances in par- 
allel, according to the rule developed above. Conversely, let sev- 
eral admittances be connected in series, and let it be required to 
find the equivalent admittance. To solve this problem, each 
parallel combination is replaced by an equivalent series combina- 
tion, and then the impedances are added in series. The student 
understands, of course, that the addition in both cases is geo- 
metric, and that only like components can be added algebraically. 
Problems of this kind occur, for instance, in the theory of trans- 
mission lines, transformers, and induction motors; for this reason 
it is important that the student understand the equivalent com- 
binations, and that he acquire facility in changing from a series 
to a parallel combination, and vice versa, as is explained in the 
preceding article. 

Prob. 1. The load of a single-phase, 6600-volt generator is estimated 
to consist of 1200 kw. of lamps, practically non-inductive, and of 800 kw. 
of motors, working at an average power-factor of 75 per cent. What 
will be the expected generator output, in amperes, and the power-factor? 
Solution: The energy component of the motor current is 800/6.6 = 121.2 
amp.; the reactive component is 121.2 tan <f> = 106.8 amp. The lamp 
current is 1200/6.6 = 181.8 amp. The total energy component of the 
generator current is 121.2 + 181.8 = 303 amp. Consequently, the total 
generator current is (303 2 + 106. 8 2 )" 2 " = 321.3 amp.; the power-factor 
is 303/321.3 = 94.3 per cent. 

Prob. 2. Check the solution of the preceding problem graphically. 

Prob. 3. Three resistances of 2, 5 and 10 ohms, and two reactances of 
4 and 2.5 ohms, are all connected in parallel across a 250- volt alternating- 
current line. What are the total current and the power-factor of the com- 
bination? Ans. 258 amp. ; 77.5 per cent. 

Prob. 4. Three impedance coils, having ohmic resistances of 2, 3 and 
4 ohms respectively, and inductances of 13, 10 and 22 millihenrys, are 
connected in parallel across a source of 220-volt, 60-cycle alternating 
voltage. Calculate the total current and the power-factor. Check the 
solution graphically. Ans. 110 amp.; cos = 0.495. 

Prob. 5. Solve the preceding problem for a frequency of 25 cycles per 
second. Construct the vector diagrams of the currents in both problems 
to the same scale, so as to see the influence of the frequency. 



Chap. VII] SUSCEPTANCE AND ADMITTANCE 81 

Prob. 6. In problem 4, let the total current be given in magnitude, 
but not in its phase position; assume the inductance of the third coil to 
be unknown. Show how to determine analytically and graphically the 
vector of the current in the third coil, and the position of the vector of 
the total current. 

Prob. 7. The admittance of a winding is 0.2 mho; the current through 
the winding lags by 34 degrees with respect to the voltage at its terminals. 
Determine the resistance and the reactance of the winding. 

Ans. 4.145 ohms; 2.796 ohms. 

Prob. 8. A coil having a resistance of 2.3 ohms and a reactance of 
5 ohms is connected in parallel with another coil, for which r = 3 ohms 
and x = 4 ohms. Calculate the resistance and the reactance of the 
equivalent series circuit. Ans. 1.36 ohms; 2.255 ohms. 

Prob. 9. The coils given in the preceding problem are connected in 
parallel across 55 volts. Calculate the total current, its energy and reac- 
tive components, and the power-factor of the combination. 

Ans. 20.85 amp.; 10.78 amp.; 17.88 amp.; cos = 0.5165. 



CHAPTER VIII 
THE USE OF COMPLEX QUANTITIES 

29. Addition and Subtraction of Projections of Vectors. 

With the explanation given in the preceding four chapters, the stu- 
dent is enabled to handle, by means of vector diagrams, problems 
involving resistances and reactances in alternating-current cir- 
cuits. A number of problems in transmission-line calculations 
and in the theory of alternating-current machinery may be solved 
by the use of such vector diagrams. The disadvantages of the 
graphical method are: (1) Results are usually obtained which 
hold for one specific case only; an analysis of the effect of various 
factors is often difficult. (2) Some vectors may be many times 
smaller than others; for instance, the voltage drop in a transmis- 
sion line, as compared to the line voltage itself. Therefore, the 
diagram must be drawn to a very large scale, or else the results 
are not sufficiently accurate. In addition to these drawbacks, 
some engineers object to graphical methods in general, as involv- 
ing the use of drawing instruments, which may not be convenient. 
On the other hand, vector diagrams are quite convenient in 
some practical cases; moreover, they are helpful for the under- 
standing of general relations in a circuit, without reference to 
particular numerical values. Again, in some problems, the un- 
known vectors can be calcu- 
lated from the vector diagram 
trigonometrically, without the 
necessity of actually drawing 
it to scale. 

It is possible to treat 
vectors analytically, using 

Y 

their projections on two axes, 

Fig. 28. A vector and its projections. as in analytic geometry (Fig. 

28). A vector, such as E, 
can be defined either by its magnitude and phase angle 6, or 
by its projections e and e' upon the axes of coordinates. 

82 



Chap. VIII] THE USE OF COMPLEX QUANTITIES 83 

If E and 6 are given, the projections are calculated from the 
expressions 

e = E cos 6; 



If the projections are given, the vector itself is determined in 
magnitude and position from the equations 

E = (e 2 + e' 2 )*; (126) 

tan 6 = e'/e (127) 

In numerical computations it is more convenient first to calculate 
tan 6 from eq. (127) and then to determine E from one of the 
eqs. (125), using trigonometric tables. This does away with the 
necessity for squaring the- projections and extracting a square 
root. 

The fact that e and e' are components of the vector E along 
two perpendicular axes is expressed symbolically thus: 

E = e+je'. ...... (128) 

Here j is a symbol which indicates that the projection e' refers 
to the vertical axis. This symbol must not have any real value; 
for the time being, it may be considered merely as an abbrevia- 
tion of the words " along the vertical axis." The sign plus in 
eq. (128) denotes the geometric addition. The dot under E sig- 
nifies that by E is meant not only the magnitude of the vector, 
but its direction as well, the latter being defined by the projec- 
tions. When the magnitude only is meant, the dot is omitted. 

The foregoing notation has been introduced by Dr. Charles 
P. Steinmetz, and is now universally used in this country. Much 
credit is also due to Dr. Steinmetz for developing the analytic 
method, used below, of dealing with alternating currents and 
voltages by means of their projections. 

The addition and subtraction of vectors are reduced simply to 
the addition and subtraction of projections. According to Fig. 12, 
the projection of a vector on any axis is equal to the sum of the 
corresponding projections of its component vectors on the same 
axis. Thus, if a current is represented as a vector by its projec- 
tions 50 + j 70 amp., and another current by 100 + j 40 amp., 
the vector sum of these currents is 150 + j 110 amp. Or, the 
resultant of two voltages, E x = e x -f je/ and E 2 = e 2 + je 2 , is 

E eq = E 1 + E 2 = ( ei + e 2 ) + j fo' + e 2 '). 



84 THE ELECTRIC CIRCUIT [Abt. 30 

As an illustration, let us solve problem 3, Art. 14, by the method 
of projections. Take the voltage vector of the first alternator 
in the horizontal direction, this being the simplest assumption. 
This vector is therefore expressed as E 1 = 2300 + j 0. The hor- 
izontal projection of the second vector is 1800 cos 27° = 1603.8 
volts, and its vertical projection is 1800 sin 27° = 817.2 volts. 
Both of these projections are positive, because the second vector 
leads the first, and is therefore in the first quadrant. Thus, E 2 = 
1603.8 + j 817.2 volts. The resultant voltage, E eq = E 1 .+ E 2 = 
3903.8 + j 817.2 volts. For some purposes, it is sufficient to 
leave the answer in this form; if, however, the magnitude and 
phase position are required, they are found as explained above: 
tan 6 = 817.2/3903.8 = 0.2092; d = 11° 49'; cos d = 0.9786; E eq 
= 3903.8/0.9786 = 3988 volts. 

If the terminals of the second machine be reversed, then E eq = 
E l — E 2 = 696.2 — j 817.2. This vector has a positive horizon- 
tal projection and a negative vertical projection. Consequently, 
it lies in the fourth quadrant, and lags behind the reference vec- 
tor by less than 90 degrees. Proceeding as above, we find = 
-49° 32'; E eq = 1074 volts. 

Prob. 1. Solve problem 1, Art. 14, by the method of projections, 
assuming the vector of the first current to be horizontal. 

Prob. 2. Check the solution of problem 4, Art. 14, by the method of 
projections. 

30. Rotation of Vectors by Ninety Degrees. In problems 
involving reactance, it is necessary to multiply the vector of the 
current by the reactance of the circuit and then turn it by 90 
degrees, in order to determine the reactive drop in voltage. The 
simple multiplication of the vector of current by the reactance 
converts it into a vector of voltage, and thus merely changes the 
scale. But turning the vector modifies the relative magnitudes 
of its projections; it is, therefore, necessary to find a relation 
between the magnitudes of the original and the new projections. 

In the simplest case let a vector E x be drawn along the refer- 
ence axis, or axis of abscissae, and let its length be a. In the sym- 
bolic notation it is represented as E x = a, the other projection 
being zero. After having been turned by 90 degrees counter-clock- 
wise, the vector is directed along the positive axis of ordinates, 
and is symbolically represented as E 2 = ja, the horizontal pro- 



Chap. VIII] THE USE OF COMPLEX QUANTITIES 85 

jection being zero. Thus, in this particular case, a rotation by 
90 degrees is equivalent to a multiplication by j. 

It is convenient to define j in such a manner that multiplication 
of any vector by j will turn the vector by 90 degrees in the positive 
direction (counter-clockwise), while division by j will turn the vec- 
tor by 90 degrees in the negative direction (clockwise) . In order to 
find a value of j which satisfies these requirements, let the vector 
E 2 be turned again by 90 degrees counter-clockwise, being now 
directed along the negative X-axis. Its expression is now E 3 = 
—a. On the other hand, the same expression must be obtained 
by multiplying E 2 by j. Therefore, we have — a = fa, or f = 
— 1; consequently, j =v— 1. If the original vector E Y is to be 
turned by 90 degrees clockwise, we must, according to our assump- 
tion, divide it by j. We then have E A = a/j, or, multiplying the 
numerator and the denominator by j, E 4 = ja/f. If j 2 = — 1, as 
it appears to be above, then E^ = —ja. This checks with the 
preceding result, because E 2 = —E 4 . It will thus be seen that 
the value of j 2 = — 1 satisfies the requirements set above, when 
the original vector is directed along one of the axes of coordinates. 

Let now the original vector Ei (Fig. 29) have an arbitrary 
direction in the first quadrant, or E x = a + jb. Multiplying E x 
by j we must get the vector E 2 , of the 
same magnitude, but in the second 
quadrant and perpendicular to E x . *—£ 
E 2 has a vertical projection equal to \ 
the horizontal projection a of the ^\ 
original vector E 1 ; the horizontal pro- 
jection of E 2 is negative, and is equal 
in its absolute value to the. vertical 
projection b of the vector E x . Thus, FlG - 29 " The relation between 
,t . , the projections of two vectors, 

the new vector is expressed as per £ endicular to each other . 
E 2 = — b + ja. On the other hand, 

multiplying E ± by j we have jEi = ja + j 2 b = ja — b, which is 
the same as above. Therefore, in this case also the assump- 
tion j 2 = — 1 is correct, and leads to rotation by 90 degrees. 
It is left to the student to verify the cases in which the vector lies 
in some other quadrant, and where Ei is divided by j, for rotation 
by 90 degrees in the negative direction. 

Expressions of the form a + jb, where a and b are real quan- 
tities and j = V— I, are called in algebra complex quantities. 



86 THE ELECTRIC CIRCUIT [Art. 30 

The student need not be discouraged by the name, because for 
our purposes j is simply a quantity which separates the two pro- 
jections of a vector, obeys the law of multiplication and division, 
and is of such a nature that j 2 =— 1. Moreover, solutions by 
means of complex quantities are quite as simple as by other 
methods. 

Prob. 1. A current of 80 + j 43 amp. flows through a resistance of 
2 ohms in series with a reactance of 3 ohms. Find the voltage drop across 
the impedance. Solution: The vector of the voltage consists of two 
components, representing the ohmic and the reactive drop respectively. 
The ohmic drop, E h is equal to 2 (80 + j 43) = 160 + j 86 volts. To 
find the inductive drop, E 2 , the vector of the current must be multiplied 
by x = 3, and then turned by 90 degrees, in other words, multiplied by j. 
Thus, E 2 = 3j(80+j43) =-129 +j 240 volts. The total voltage 
E = E x + E 2 = 31 + j 326 volts. 

Prob. 2. Solve the preceding problem when the voltage is given and 
the current is unknown. First Solution: Let the unknown current be 
represented by its projections as i + ji'. We have, as in the preceding 
problem, 

2 (i+ji') + Sj(i+m = 3I+J326, . . . (129) 

or, collecting the terms containing j, 

i2i-3i')+j(2i' + 3i) = 31+^326. . . . (130) 

This equation can be satisfied only if the terms with and without j 
are equal to each other respectively, because a real quantity cannot be 
equal to an imaginary one. Or, from a geometric point of view, the left- 
hand side and the right-hand side of eq. (130) each represent a vector by 
its projections. But two vectors are identical only when their corre- 
sponding projections are equal. Thus, we have 

2i-3i' = 31; 2i' + 3^ = 326. 

Solving these equations for i and i' , we find i = 80, i' = 43, as in the pre- 
ceding problem. Second Solution: Equation (129) can be written in the 
form (2 + 3 j) (i + ji') = 31 + J326; or, i+ ji'= (31 + j326)/(2 +3j). 
Considering here j as an ordinary algebraic quantity, we can get rid of 
it in the denominator by multiplying both the numerator and the denomi- 
nator by 2 — 3 j. The result is 

.., (31+j326)(2-3j) _1040 .559 
1_h ^ " 2'-(3j> " 13 ^ J 13' 

or i + ji' = 80 + j 43, as before. 

Prob. 3. A voltage of 28 + j 120 volts applied to the terminals of a 
coil produces in it a current equal to 4 + j 1.5 amp. Determine the 
resistance and the reactance of the coil. 

Ans. r = 16 ohms; x = 24 ohms. 



Chap. VIII] THE USE OF COMPLEX QUANTITIES 87 

Prob. 4. Verify the answer to problem 4, Art. 28, by the method of 
projections, assuming the vector of the voltage to be horizontal. 

31. Impedance and Admittance Expressed as Complex Quan- 
tities or Operators. Let it be required to find the voltage neces- 
sary to maintain a current i -f ji' through a resistance r and react- 
ance x in series. The voltage drop in the resistance is r (i -\-ji')', 
that in the reactance is jx (i + ji') . Hence, the total voltage is 
E = r (i + ji') + jx (i + ji'), or 

E = e + je' = (r + jx) (i + ji'). . . . (131) 
It is legitimate to factor out the expression (i + ji'), and to treat 
j as any other algebraic quantity, because j is now assigned a 
definite value, V — 1. Moreover, eq. (131) represents simply the 
geometric addition of four component vectors, two of them 
directed along the X-axis and the other two along the F-axis. 
As long as this interpretation is kept in mind, the terms may be 
arranged in any desired order. 

Equation (131) shows that, in order to obtain the expression 
of the voltage drop through an impedance, the current must be 
multiplied by the complex quantity r + jx. The expression 
r + jx is not a vector, because it does not stand for a sine-wave, 
but an operator upon the vector of the current. The operation 
consists, first, in multiplying the vector of the current by r, then 
in multiplying the same vector by x and turning it by 90 degrees 
in the positive direction, and finally, in adding the two vectors 
geometrically. All these operations are included in the expression 
r -f- jx, which is called the impedance operator. 

In order to get the projections e and e' of E from eq. (131), 
the terms on the right-hand side must be actually multiplied and 
the results represented in the form of a complex quantity. We 
get then, separating the real and the imaginary parts, 
E = e + je' = (ri — xi') + j (ri' + xi) . 

The real and the imaginary parts on each side of this equation are 
equal to each other respectively, because they represent the pro- 
jections of the same vector E upon the two axes. Consequently 

e = ri — xi' 



(132) 
r = rv + xi. ' 

In problems these steps are best left until the numerical values 
have been substituted, in order to avoid complicated expressions. 



88 THE ELECTRIC CIRCUIT [Art. 31 

If the voltage and the impedance are given, and it is required 
to find the current, we get from eq. (131) the relation 

i + ji' = (e + je')/(r + jx) . 
In order to reduce the right-hand side of this equation to the 
form of a complex quantity, we multiply the numerator and de- 
nominator by the expression r — jx. This gives 

, _ (e+je')(r - jx) _ re + xe' . re' - xe 
l ~ tJl ~ r*-{jxf ""r 2 +x 2+J r 2 + x 2, U } 

Equating the real and the imaginary parts respectively, we 
obtain 

i = (re + xe')/z 2 ; 1 

i' = (re' - xe)/z\ j {i6V 

Equation (131) expresses the fact that the voltage E is equal 
to the product of the current by the impedance, if the operator 
(r + jx) be considered as the impedance of the circuit in the com- 
plex notation. Denote the impedance by Z, then 

Z = r+jx. ....... (135) 

Here capital Z is used to indicate that* it is a complex quantity, 
as distinguished from the numerical value z of the same imped- 
ance. The letter is not provided with a dot, because Z is not a 
vector, but an operator. 

In the abbreviated notation, eq. (131) becomes 

E = IZ (136) 

In this expression each letter stands for a complex quantity, so 
that when actual numerical or algebraic relations are necessary, 
the expression must again be expanded into (131) and the multi- 
plication of the two complex quantities actually performed. 

Instead of dividing the voltage by the operator (r -j- jx) and 
then eliminating j from the denominator, it is more convenient 
to introduce another operator by which the voltage must be multi- 
plied in order to obtain the current. It will be remembered from 
Art. 26, that a voltage must be multiplied by an admittance in 
order to get the current. Consequently the operator in question 
must be expected to have the elements and the dimensions of an 
admittance. Replacing the given series combination by an equiv- 
alent parallel combination (Art. 27), the unknown current is split 
into a component Eg through a pure conductance, and a compo- 
nent — jEb through a pure susceptance in parallel with the con- 



Chap. VIII] THE USE OF COMPLEX QUANTITIES 89 

ductance. The latter component is provided with the prefix —j, 
because it lags by 90 degrees behind the voltage. Thus, the total 

current 

!=E(g-jb) (137) 

The expression 

Y = g-jb (138) 

is called the admittance operator. The symbol Y, like the symbol 
Z above, is not provided with a dot because it is not a vector. 
Combining the two preceding equations gives 

I = YE (139) 

Equations (136) and (139) represent generalized forms of Ohm's 
law for alternating currents, corresponding to the simple expres- 
sions (1) and (2) for direct current. Equation (139) is an abbre- 
viated form of the relation 

i+ji' = (e + je')(g-jb) (140) 

Multiplying out and equating the real and the imaginary parts 
on both sides of this equation, we get 

i = ge + be'; 



141 
% = — be + ge . ' 

The relations between r, x and z on one hand, and g, b and y on 
the other, are deduced in Art. 27; it being understood, of course, 
that in the present treatment the two combinations are equivalent. 
Equations (136) and (139) imply that 

YZ = 1, (142) 

or 

(r+jx) (g -jb) = 1. 

Substituting into this last equation the values of g and b from 
eqs. (121) and (122) and performing the multiplication, it will be 
found that the equation is reduced to the identity 1 = 1, this 
being a check on eq. (142). 

With the abbreviated notation of complex quantities, using the 
symbols 1?, I, Z and Y, alternating-current problems are solved 
almost as easily as direct-current problems. Either the impedance 
operator or the admittance operator is used, depending upon the 
relative connection of the parts of the circuit, whether parallel 
or series. In many cases the abbreviated notation may be pre- 
served until the solution has been obtained, the projections of 
the vectors, e + je f and i + ji' , and the expanded forms of the 



90 THE ELECTRIC CIRCUIT [Art. 31 

operators, r + jx and g—jb, only then being substituted in a 
numerical form to get the final answer. 

Prob. 1. Find, by means of complex quantities, the voltage required 
in problem 5, Art. 21. Solution: Assume the vector of the current to 
be the reference vector. At the power house cos <t> = 7520/(66 X 147) = 
0.775; sin0 = 0.632. The generator voltage is 66 X 0.775 + j66 X 
0.632 = 51.15 +j 41.71 kilovolts. The voltage drop in the line is 
147(45 +j 83) = 6615 + j 12,200 volts. Hence, the load voltage is 
(51.15 - 6.61) + j (41.71 - 12.2) = 44.54 + j 29.51 kilovolts, The mi- 

/ 2 2 \ — 

merical value of the load voltage is (44.54 + 29.51 J 2 = 53.43 kilo- 
volts. 

Prob. 2. Determine analytically the resistance r 2 required in problem 
2, Art. 22. 

Prob. 3. A voltage equal to 180 + j 75 produces a current of 7 + 
j 1.5 amp. What is the impedance of the circuit? 

Ans. 26.78 + j 4.97 ohms. 

Prob. 4. Power is transmitted from a single-phase alternator to a 
load consisting of a resistance of 1.17 ohms in series with a reactance of 
0.67 ohm. The generator voltage is 2300, and the impedance of the 
transmission line is 0.085 + j 0.013 ohm. Determine (a) the line cur- 
rent; (b) the voltage drop in the line; (c) the receiver voltage. Take 
the generator voltage as the reference vector. 

Ans. (a) 1413.6- j 769.6 amp.; (b) 130.1 -j 47 volts; (c) 2169.9 + 
j 47 volts. Use the admittance operator to obtain the current, and the 
impedance operator to calculate the line drop. 

Prob. 5. A voltage, e + je', is impressed across the impedances 
ri + jxi and r-i -f- jx* in parallel. Find the total current. Solution: 
The total conductance is g = n/21 2 + r 2 /2 2 2 , and the total susceptance is 
b = Xi/zi 2 + X2/Z2 2 . Hence, the current i + ji' = (e + je') (g — jb) = 
(eg+e'by + j&g-eb). 

Prob. 6. Extend the solution of the preceding problem to the case 
in which more than two impedances are in parallel. 

Ans. i + ji' = [e2(r/2 2 ) + e'z(x/z 2 )] + j [e'x(r/z 2 ) -.eX{xfz% 

Prob. 7. Two impedances, n + jx x and r 2 + jx 2 , in parallel, are 
connected in series with a third impedance r + jx. Show how to deter- 
mine the total voltage, knowing the total current i + ji' ; or, how to find 
the expression for the total current when the total voltage e + je' is 
given. 

Prob. 8. Show how to solve the preceding problem when both the 
current and the voltage are given, but either the impedance n + jx x or 
the impedance r + jx is unknown. 



CHAPTER IX 

THE USE OF COMPLEX QUANTITIES — (Continued) 

32. Power and Phase Displacement Expressed by Projec- 
tions of Vectors. Let an alternator supply a current I = i -\- ji' 
at a voltage E = e -\- je' ', and let it be required to calculate the 
power output of the generator. The expression for the average 
power is P = EI cos </>, where </> is the phase displacement between 
E and I. The angle <£ is the difference between the angles 6 e and 
6i which the vectors E and / respectively form with the reference 
axis. Hence, we have 

P = EI cos </> = EI cos (0 e - Oi) 

= E cos 6 e • / cos di + E sin e • I sin 0». 

Remembering that E cos e , E sin e , etc., represent the projec- 
tions of the given vectors on the axes of coordinates, we have 
simply 

P = ei + e'i' (143) 

Another way of deducing expression (143) is to resolve the given 
vectors of current and voltage into their components along the 
axes of coordinates, and to consider the contribution of each pro- 
jection to the total power. The projections e and i, being in 
phase, give the power ei. Similarly, the projections e' and i' give 
the power e'i' . The projection i' of the current gives zero average 
power with the projection e of the voltage, the two being in phase 
quadrature. For the same reason the average power resulting 
from e' and i is equal to zero. Thus, ei + e'i' represents the 
total average power. 

To find the phase displacement, or the power-factor of the 
output, we write 

, " . . tan e — tan 0< 

tan = tan (0 e - 0») = -— - — r— — - , 
1 + tan B e tan 0» 

or 

ton^ («V«)-(*'A) (1U) 

tan * - i+(e'A0-(i'A)' (1 } 

Knowing tan 0, its cosine is found from trigonometric tables. 1 

1 Or else the power-factor, cos = cos (d e — t ), can be found from the 
relations d e = tan -1 e'/e, and 6 t = tan -1 i'Ji. 

91 



92 THE ELECTRIC CIRCUIT [Art. 32 

Power-factor can also be determined directly from the expres- 
sion 

cos = P/EI = (ei + eV)/[(e 2 + e' 2 ) (i 2 + i' 2 )}^, (145) 

but the calculations are more involved than when formula (144) 
is used. 

The power calculated by means of formula (143) sometimes 
comes out negative, if some of the projections of E and / are 
negative. The interpretation is that the phase displacement 
between the current and the voltage is over 90 degrees, so that 
power is being supplied to the machine, instead of being delivered 
by it. In other words, the machine acts as a motor and not as a 
generator. Tan in formula (144) may also be negative, which 
means either that the current is leading, or that it is lagging by 
an angle larger than 90 degrees. The question is decided by 
reference to the sign of the power. 

For the reactive power (Art. 19) we have 
P r = EI sin = EI sin (6 e - Oi) = E sin d e I cos 0,- - E cos B e I sin if 

or P r = e'i - ei' (146) 

The apparent power is 

P a = (i 2 + i ,2 f (e 2 + e' 2 f (147) 

However, it is sometimes more convenient to determine the appar- 
ent power from the relation 

P a =P/cos0, ..;... (148) 

where P is calculated from eq. (143), and cos is found from 
trigonometric tables, knowing tan from eq. (144). 

Prpb. 1. The terminal voltage of an alternator is 5370 + J735; the 
line current is 173 — j 47 amp. Calculate the output of the machine and 
the power-factor of the load. Ans. 894.5 kilowatts; 92 per cent. 

Prob. 2. In the preceding problem, what must be the projection of the 
current upon the F-axis in order that the power shall become zero? 

Ans. i'= —1264 amp. 

Prob. 3. Let the line current in problem 1 be — 58+J12 amp. 
Explain the negative sign of the power and the plus sign of tan </>. Draw 
the vectors of the current and voltage. 

Prob. 4. A synchronous machine generates a voltage equal to 2300 — 
j 50 volts, and supplies a current, through an impedance of 5 + j 50 
ohms, to another synchronous machine generating a counter-e.m.f. of 
2300 + j 50 volts. What is the power output of the first machine? Is 
the current leading or lagging? Make clear to yourself the physical 
meaning of the answer. . Ans. —4.55 kw.; <t> = 173° 3' lagging. 



Chap. IX] THE USE OF COMPLEX QUANTITIES 93 

Prob. 5. A current of 350 — j 75 amp. is maintained, through an im- 
pedance, the power output being 952 kw. at a power-factor of 86 per cent 
lagging. Find the voltage across the impedance. Hint: Solve eqs. (143) 
and (144) together for the unknown projections e and e' '. 

Ans. 2930 + j 987 volts. 

Prob. 6. Solve problem 5 by calculating the value of the impedance, 
and multiplying the impedance by the current. Hint: power = I-r; 
x = r tan <t>. 

Prob. 7. Solve problem 5, using the expression for the reactive power. 

33. Vectors and Operators in Polar Coordinates. Instead of 
representing a vector by its orthogonal projections, as in eq. (128), 
it is sometimes more convenient to express the same vector as a 
complex quantity in terms of its magnitude and direction. Sub- 
stituting the values of e and e' from eqs. (125) into eq. (128), we 
obtain 

E = E (cos d + j sin 6) (149) 

Similarly, a current in phase with this voltage is expressed as 

1=1 (cos d + j sin B), (150) 

while a current lagging by an angle behind the voltage E is 
represented by the equation 

/= 7[cos(0- 0) +jsin(0 - 0)]. . . . (151) 

When the vectors of currents and voltages are expressed in the 
trigonometric form shown above, it is convenient to use the 
operators Z and Y in a similar form. Substituting the values of 
r and x from eqs. (95) and (96) into eq. (135), we get 

Z = z (cos + i sin 0) (152) 

In a similar manner, using eqs. (116) in eq. (138), gives 

F = y (cos 4> — j sin <£) (153) 

When calculating the voltage drop IZ or the current E/Z it 
is necessary to find the product or the ratio of two complex 
expressions of the form cos 6 + j sin 6. By actually performing 
the multiplication and separating the real from the imaginary 
term we find that 

(cos0+j sin0) (cos0+jsin0)=cos {9+4>)-\- j sin (0+0). (154) 

This gives a simple rule for the multiplication of two or more 
complex quantities in the trigonometric form. In order to deduce 



94 THE ELECTRIC CIRCUIT [Art. 33 

a similar rule for division, we observe that 

l/(cos 4> + j sin 0) = 

cos0 — j sin0 = cos ( — 0) +./ sin(— 0). . . (155) 

This relation is easily verified by multiplying the numerator and 
the denominator of the left-hand side of the equation by cos — 
j sin 0, so as to get rid of the complex quantity in the denomi- 
nator. Equation (155) leads to the following rule for the division 
of complex quantities in trigonometric form : 

(cos 0-f-jsin 0)/(cos0+,7sm0)=cos (0 — 0) + j sin (0 — 0). (156) 

Thus, for instance, if the current given by eq. (150) flows through 
an impedance expressed by eq. (152), the required terminal volt- 
age is 

E = IZ = Iz [cos (0 + 0) + j sin (0 + 0)], . (157) 

which result simply means that the voltage is equal to Iz and 
leads the current by the angle 0. 

The operator given by eq. (152) multiplies a vector by z and 
turns it by the angle in the positive direction. Hence, the 
operator (cos + j sin 0) simply turns a vector by the angle 0, 
without changing its length. Thus, if it be required to turn a 
vector A= a + ja' by an angle a in the positive direction, the 
projections of the new vector are found from the following expres- 
sion: 

(a + ja f ) (cos a + j sin a) — (a cos a — a! sin a, 

-j- j (a f cos a + a sin a) (158) 

Of course, the same result could be obtained by first calculating 
the angle which the vector A forms with the reference axis, from 
the relation tan = a' /a, and then determining the new projec- 
tions A cos (0 + a) and A sin (0 + a). 

Voltage Regulation of a Transmission Line. 1 As an example of 
the use of complex quantities in the trigonometric form, let us 
consider the voltage regulation of a single-phase transmission 
line. Let the resistance and reactance of the line, and the gen- 
erator voltage Ei, be given; and let it be required to determine 
the receiver voltage E 2 for a given current I and a given power- 

1 The electrostatic capacity of the line is disregarded here; a complete 
treatment of the regulation of a transmission line, taking into account the 
capacity and leakage, is given in Arts. 68 and 69 at the end of the book. 



Chap. IX] THE USE OF COMPLEX QUANTITIES 95 

factor of the load cos 0'. In the symbolic notation we have 

E x = E 2 + IZ, (159) 

where the impedance Z of the line is known, and is expressed by 
eq. (152). The phase angle refers to the line, the angle 0' to 
the load. 

When actually solving an equation such as (159), it is highly 
important to select the reference axis in the most advantageous 
way, so as to simplify the calculations as much as possible. In 
the case under consideration, it is convenient to select the refer- 
ence axis in the direction of E 2 , because then E 2 is determined 
by its magnitude alone, the direction angle being equal to zero. 
The generator voltage is expressed by Ex (cos 6 + j sin 6), where 
the magnitude of Ex is given, but the angle 6 is unknown. The 
current lags by the angle 0' behind E 2 , and therefore is expressed 
by the formula I (cos </>' — j sin </>'). Thus eq. (159) becomes 

#i (cos 6+j sin 6) = E 2 + Iz [cos (0 - 0') + j sin (0 - 0')]- (160) 

Equating the real and the imaginary parts gives 

Ei cos = E 2 + Iz cos (0 - 0') ; • • . (161) 

Ex sin 6 = Iz sin (0 - 0') (162) 

From eq. (162) 

sin 6 = (Iz/Ex) sin (0 - 0') (163) 

Knowing 0, we find from eq. (161) 

E 2 = Ex cos 6 - Iz cos (0 - 0'). . . . (164) 

In practice, one is usually required to determine the voltage 
regulation of the line. According to the definition adopted by 
the American Institute of Electrical Engineers {Standardization 
Rules, Art. 187), 

per cent regulation = 100 (E 2 - E 2 )/E 2 , . . (165) 

where E 2 is the value of E 2 at no load. But here E 2 = E 1} be- 
cause the electrostatic capacity of the line is neglected. It is pos- 
sible to determine the difference Ex — E 2 directly from eq. (161), 
by substituting for cos 6 the expression 1 — 2 sin 2 \ 6. We obtain 
then 

AE = Ex- E 2 = Iz cos (0 -00+2 Ex sin 2 \ 0. (166) 

Equation (165) becomes 

per cent regulation = 100 AE/(Ex - AE). . (167) 



96 



THE ELECTRIC CIRCUIT 



[Art. 33 



When it is required to calculate the voltage regulation for several 
loads, the computations are conveniently arranged in a table 
of the following form: 



Load 
current 

/ 


Line 

drop 

Iz 


Load 
power- 
factor 

COS 0' 


Angle 

0-0' 


sin (0—0') 


COS (0 — 0') 


sin0 


Angle 
d 


sin 2 i d 


AE 


Per cent 
regula- 
tion 

























In practice, the voltage regulation is usually required for a 
certain load of P 2 watts, so that, strictly speaking, the current / 
is not known. But, since E 2 is not much different from E 1} it 
is an easy matter to estimate the current with sufficient accuracy. 
Or else a curve of voltage regulation is plotted against the load 
as abscissa?, so that the regulation may be read off at any desired 
load. It is possible to solve the problem exactly, by using for 
/ its expression P 2 /(E 2 cos 4> f ) in eqs. (161) and (162). In this 
case the equations are squared and added, so as to eliminate 0. 
This gives a biquadratic equation for E 2 , from which the receiver 
voltage can be computed. 

This problem can also be solved when the complex quantities 
are expressed in the orthogonal form, instead of the trigonometric 
form here used. The student is urged to work out the details, in 
order to become thoroughly familiar with complex quantities in 
both forms. 

Prob. 1. A vector 72 -\- j 53 must be turned by 25 degrees in the nega- 
tive direction. What are its new projections? Ans. 87.65 + j 17.6. 

Prob. 2. A single-phase aluminum line is to be built from a power 
house, at which a voltage of 11,500 is maintained at a frequency of 50 
cycles, to a point 25 km. distant. When a current of 60 amp. at 80 per 
cent lagging power-factor is delivered at the receiver end, the power loss 
in the line must not exceed 10 per cent of the useful power. What must 
be the size of the conductor, and what will be the per cent voltage regula- 
tion at this load? The spacing between the wires is to be 61 cm. 

Ans. No. 0000 B. & S.; 11.4 per cent. 

Prob. 3. Check the answer to the preceding problem graphically. 



Cuap. IX] THE USE OF COMPLEX QUANTITIES 97 

Prob. 4. Explain the theory of Mershon's diagram found in various 
electrical handbooks and pocketbooks, and check by means of it the 
answer to problem 2. 

Prob. 5. Show how to determine the voltage regulation of a trans- 
mission line when the receiver voltage is given. 

Prob. 6. Show how to calculate the receiver voltage Ei from eq. (159), 
using the orthogonal projections of the vectors and operators. Discuss 
the relative advantages and disadvantages of the rectangular and polar 
coordinates in this case. 

34. Vectors and Operators Expressed as Exponential Func- 
tions. 1 Expressions (149) to (153) are sometimes written in the 
exponential form, using the identity 

cos0+jsin0 = e#, (168) 

where e is the base of natural logarithms. This important equa- 
tion follows from the well-known expansions for sin 0, cos and 
e 6 , obtained by Maclaurin's Theorem in calculus; namely, 

3 5 
sin = 0-^ + ^- . . . ; 

cos0=l-2j+ iI - 

€ =1+ T + 2! + 3! + 4! + 

The last series, when j is substituted for 0, becomes 

2 3 4 
1 ^ J u 2! J S\ 4! ' ' ' ' 
Substituting these values into eq. (168), it is found to be an iden- 
tity. Thus, we have 

E = E (cos + j sin 0) = E^ (169) 

Similarly, the impedance operator becomes 

Z = z (cos <f>+j sin 0) = ze>*, .... (170) 
and the admittance operator 

Y = y (cos <f> —j sin 0) = ye~^. . . . (171) 
If, for instance, a current is given as / = Ie ld and if it flows through 
an impedance Z = ze i4> , the required voltage is found by multi- 
plying these two expressions, or 

E = IZ = W+O (172) 

1 This article may be omitted if desired, without impairing the conti- 
nuity of the treatment in the rest of the book. 



98 THE ELECTRIC CIRCUIT [Art. 34 

This shows that the absolute value of the voltage is Iz and that 
it leads the current by the angle 0. Equation (172) corresponds 
to eq. (157) in trigonometric notation. The projections of the 
current vector are I cos d and Zsin0; while the projections of 
the voltage vector are Iz cos (0 + 0) and Iz sin (6 + 0). Thus, 
it is always possible to change from the exponential form to the 
trigonometric form, and finally to the orthogonal projections, or 
vice versa. The exponential form is more concise, and possesses 
marked advantages in the solution of some advanced problems 
relating to alternating currents and oscillations. 1 However, for 
the simple problems treated in this book, the plain algebraic nota- 
tion a + ja f and the trigonometric notation A (cos a + j sin a) 
are amply sufficient. It is deemed advisable to explain the expo- 
nential notation here in order to enable the student to read books 
and magazine articles in which it is employed. 

1 See for instance J. J. Thomson, Recent Researches in Electricity, and 
Magnetism. 



CHAPTER X 
POLYPHASE SYSTEMS 

35. Two-phase System. The student knows from his ele- 
mentary work that the induction motor operates on the principle 
of the revolving magnetic field, and that such a field is produced 
by a combination of two or more alternating currents differing 
in phase. An electric circuit upon which are impressed two or 
more waves of e.m.f. having definite phase displacements is called 
a polyphase system. A large majority of the alternating-current 
circuits used in practice in the generation and transmission of 
electrical energy are polyphase systems; it is therefore essential 
that the student become familiar with the current and voltage 
relations in such circuits. 

Theoretically, the simplest polyphase system is a four-wire 
two-phase system (Fig. 30), although it is not the most econom- 



K Phase 1 > 




*$( 



Phase 2 



Receiver 



Fig. 30. A four-wire two-phase system with two independent circuits. 

ical one in practice. The two generator windings are independ- 
ent, and are relatively displaced by ninety electrical degrees. 
The two alternating voltages induced in these windings are there- 
fore displaced in time phase by a quarter of a cycle. Each phase 
may be used separately, for instance for lighting, or both phases 
may be combined in the windings of a synchronous or induction 
motor, for the production of a revolving magnetic field. Each 
phase can be treated separately, as if it belonged to an independ- 
ent single-phase system. 

99 



100 



THE ELECTRIC CIRCUIT 



[Art. 35 



Some economy in line conductors and insulators is achieved by- 
combining two conductors belonging to different phases into one 
return conductor (Fig. 31). Such a system is called a three-wire 

A Ii. L' 



> Generator 



E, 



B 



B' 



Receiver 



b' 



G ^7 G' 

Fig. 31. A three-wire two-phase system. 

two-phase system. The current and voltage relations for a balanced 
load and a lagging current are shown in Fig. 32. The vectors E x 

and E 2 represent the voltages 
induced in the two generator 
or transformer windings from 
the point to the points A 
and B respectively; in other 
words, they are the voltages 
between each phase wire and 
the return wire. The vector 
E12* is the geometric difference 
of the two, and represents the 
voltage between the two phase 
wires. That E X2 is the differ- 
ence and not the sum of Ei 

", of currents and E t is proved by the 

following reasoning: Let the 
wire OGG'O' be permanently 
grounded, so that its potential is zero. Let the potential of the 
wire A A' at a certain instant be for example 100 volts above 
the ground, and that of the wire BB' 60 volts above the ground. 
Then the difference of potential, or the voltage between A A' and 
BB', is 40 volts. The same reasoning applies to every instant, 
so that the vector of the voltage between A A' and BB' is the 
geometric difference between E x and E 2 , which are the vectors 
of the voltages between the points A and O, and B and O re- 
* Pronounced E — one — two, and not E sub twelve. 




and voltages for the two-phase system 
shown in Fig. 31. 



Chap. X] 



POLYPHASE SYSTEMS 



101 



spectively. That is, the voltage between A and B is represented 
in phase and magnitude by the vector connecting the ends of the 
vectors Ei and E 2 . Numerically, 

E 12 = E l V2 = E 2 V2 (173) 

The currents in the conductors A A' and BB' are represented 
by the vectors h and I 2 , lagging by an angle 4> with respect to the 
corresponding voltages. The current in the return conductor is 
the geometric sum of the two phase currents, and is represented 
by the diagonal vector I u . It will thus be seen that the common 
return current is V2 times as large as each component current, 
or 

I 12 = I 1 V2 = I 2 V2 (174) 

If it is desired to have the same current density in each of the 
three conductors, the cross-section of the return wire must be 
V2 times that of each of the other two wires. 

The two phases in Fig. 30 are sometimes electrically inter- 
connected at their middle points, as shown in Fig. 33 at the left. 




Fig. 33. A star-connected quarter-phase system to the left, a mesh-connected 
system to the right. 

This is done in order to fix the difference of potential between the 
two phases. If the voltage between A and B is E, then the volt- 
age between the common point O and each wire is J E, and the 
voltage between the two phases is equal to J E V2 = E/V2.- For 
example, the voltage E da between D and A equals E a — E d . These 
relations are shown vectorially in Fig. 34. This circuit is some- 
times called the star-connected quarter-phase system. 

The four windings of a generator or motor are sometimes con- 
nected in mesh, as indicated in Fig. 33 to the right. With the 



102 



THE ELECTRIC CIRCUIT 



[Art. 35 



star connection, the star voltages OA, OB, etc., are induced 
directly, while the mesh voltages AD, DB, etc., are established by 
the combination of the star voltages. With the mesh connection 
of the windings, however, the mesh or line voltages are induced 
directly. The mesh and star voltages are shown in Fig. 34. 
Electrically the two arrangements are equivalent, provided that 
the proper numbers of turns are used in the windings. 

The line and mesh currents are indicated in Fig. 34. The line 
currents and those in the star-connected windings are represented 




Fig. 34. A vector diagram of currents and voltages in the quarter-phase 
system shown in Fig. 33. 



by the sides of the square, each current lagging by the angle 
with respect to the corresponding star voltage; the angle of lag 
depends upon the character of the load. In Figs. 33 and 34 the 
voltages are taken in the cyclic order AC, CB, BD, DA; hence, it 
is natural to take the positive direction of the current in the same 
way. With the arrows in Fig. 33 showing the positive directions 
of the currents, each line current is the difference between two 
adjacent mesh currents. Hence, in the vector diagram the mesh 



Chap. X] 



POLYPHASE SYSTEMS 



103 



currents are represented by the radii from the center to the vertices 
of the current square. It will be seen that the angle between the 
mesh currents and the mesh voltages is also equal to 0. While 
the mesh voltages are V2 times as large as the star voltages, the 
mesh currents are 1/V2 times the star currents. This condition 
is necessary in order to have the same power per phase in the 
mesh and star-connected systems. 

36. Three-phase Y-connected System. This system is 
shown in Fig. 35; the current and voltage relations are repre- 




r 




B' 



Fig. 35. A three-phase Y- or star-connected system. 

sented in Fig. 36. OA, OB, and OC represent three generator 
windings; O'A', O f B\ and O'C are the windings of a receiving 
apparatus — for instance, an in- 
duction motor. The three gener- 
ator windings are placed on the 
armature core at angles of 120 
electrical degrees with respect to 
each other, so that the alternating 
voltages induced in these windings 
are displaced in phase by one- 
third of a cycle, the positive di- 
rection in the windings being 
outward. They are represented 
by the vectors E a) E b , and E c in 
Fig. 36. The motor windings are 
similarly displaced, so that the 

whole system is symmetrical with respect to the three phases. 
The currents lag behind the voltages by an angle 4> depending 
upon the relative amounts of resistance, reactance, and counter- 
e.m.f. in the circuit. 




Fig. 36. The line and star voltages, 
and the line currents in the Y- 
connected system shown in Fig. 35. 



104 THE ELECTRIC CIRCUIT [Art. 36 

The diagram of connections shown in Fig. 35 is also called the 
star connection, and the points and 0' are called the neutral 
points of the system. The voltages between the line conductors 
and the neutral points are called the star or phase voltages, as 
distinguished from the line voltages, or voltages between any 
two line conductors. A line voltage, for instance between the 
points A and B, is equal to the geometric difference between the 
voltages OA and OB, as has been shown above in the case of a 
two-phase line. Consequently, the line voltages are represented 
in Fig. 36 by the three vectors E a b, E bc , and E ca , which connect 
the ends of the vectors of the phase voltages. It will be seen that 
the line voltages are Vs times as large as the phase voltages. 

When the three phases are perfectly balanced and the currents 
are nearly sinusoidal, the two neutral points and 0' may be 
connected by a wire, as shown by the dotted line, or grounded, 
and very little current will flow through this connection. The 
reason is that the algebraic sum of the three currents flowing 
towards or from the neutral points is equal to zero at all instants, 
because 

sin u + sin (u + f n) + sin (u — f it) = 0. (175) 

This identity is easily proved by expanding the left-hand member, 
using the expression for the sine of the sum of two angles. It 
will also be seen from Fig. 36 that the geometric sum of the three 
current vectors is equal to zero, because these vectors when added 
form a closed triangle. In practice, there are transmission lines 
on which one or both neutrals are grounded, although in some 
installations both neutrals are insulated from the ground. To 
prevent large currents with unbalanced loads or during short- 
circuits, the neutrals are often grounded through protective 
resistances. The question of grounded vs. ungrounded neutrals is 
still in a somewhat controversial stage. 

The power developed in the generator windings and available 
at the generator terminals is 3 IyEy cos <f>, where Ey is the phase 
voltage. Since the line voltage E A = Ey V^3, we have 

P = 3 I y Ey cos = I Y E A Vd cos 0. . . (176) 

In practical calculations of three-phase transmission lines and 
electrical machinery, only one phase is considered; that is, the 
three-phase circuit is reduced to an equivalent single-phase cir- 
cuit. Let it be required, for example, to calculate the cross-sec- 



Chap. X] , POLYPHASE SYSTEMS 105 

tion and per cent voltage regulation of a three-phase 66,000-volt 
line, to transmit 50,000 kw. at 80 per cent power-factor, and at 
a loss of 10 per cent of the useful power; the spacing to be 1.8 m. 
First of all we find that the voltage between each wire and the 
neutral is 66,000/ V3 = 38,100 volts, and that the power per 
phase is 50,000/3 = 16,700 kw. Hence, the problem is reduced 
to the following one: Determine the cross-section and per cent 
voltage regulation of a single-phase 38,100-volt line, having a 
spacing of 1.8 m., the i 2 r loss in one conductor being 1670 kw., 
and the resistance of the return conductor being negligible. The 
solution of this problem is given in Art. 33 above. A drop of say 
5 per cent in the phase or star voltage means also a drop of 
5 per cent in the line voltage, because of the fixed ratio 1/Vs 
between the two. 

Prob. 1. Assuming the reference axis in Fig. 36 to be horizontal, the 
line voltage equal to 44 k v., the current per phase 73 amp., and the angle <f> 
equal to 15 degrees, write down the complex expressions for all the cur- 
rents and voltages. 

Ans. E b = 22-jl2.7kv.;E ab =22 -j38.1kv.;/ a = 18.9 +j 70.5 
amp. 

Prob. 2. A three-phase 60-cycle line is 16 km. long; the spacing be- 
tween the wires is symmetrical and is equal to 61 cm., the conductors con- 
sisting of copper wire of 14 mm. diameter. It is required to maintain 
a voltage of 6700 between the conductors at the receiver end of the line. 
What is the generator voltage when the load is equal to 1000 kw. at unity 
power-factor? Ans. 7040. 

Prob. 3. Show that a three-phase transmission line may be treated 
as a single-phase line which transmits one-half the power at the same 
voltage. The three-phase line requires three conductors of the same size 
as the single-phase line, with the same spacing (25 per cent saving in 
material). 

Prob. 4. When the phase currents have higher harmonics, show that 
equalization currents must flow through the neutral connection, even 
though the phases are perfectly balanced. What happens when the 
neutrals are insulated from each other? 

Prob. 5. Show that the line voltage cannot have the third, the ninth, 
the fifteenth, etc., harmonics, even if these harmonics are present in the 
phase voltages. 

37. Three-phase Delta-connected System. This method of 
three-phase connection is shown in Fig. 37, one end of the line 
being connected, for instance, to an alternator, the other end to a 
motor or to three transformers. Fig. 38 represents the current 



106 



THE ELECTRIC CIRCUIT 



[Art. 37 



and voltage relations. The currents in the windings are differ- 
ent from those in the line. With the positive directions of the 




;^l_ 




Fig. 37. A three-phase delta- or mesh-connected system. 

currents indicated in Fig. 37, each line current is equal to the 
difference between the two adjacent currents in the " delta." 




/0/ 



Fig. 38. The voltages and currents in the delta-connected system 
shown in Fig. 37. 

Hence, in the vector diagram the line or " star " currents are 
represented by a triangle, and the delta currents by the rays from 
the center to the vertices of the triangle. It will be seen that the 
delta currents are equal to 1/Vs of the line currents, and are 
displaced in phase by 30 degrees with respect to them. This also 
follows from the identity 

J sin u - I sin (u + 120°) = V3 / sin (u - 30°). . (177) 



Chap. X] POLYPHASE SYSTEMS 107 

While there are no neutral points with a delta system, one or more 
of them may be artificially created by connecting three resistances 
or reactances in star as shown in Fig. 37 by dotted lines. The 
Y-voltages between this neutral and the line conductors are 
shown in Fig. 38 by the vectors E a , Eb and E c . The delta volt- 
ages are Vs times as large as the star voltages, while the star or 
line currents bear the same ratio to the delta currents. This is 
necessary in view of the power relation 

P = 3 EJ A cos <f> = 3 E y Iy cos 0. . . . (178) 

In design and performance calculations one phase only is 
considered, the three phases being identical when the load is 
balanced. As far as the line is concerned, the delta-connected 
generator and load may be replaced by equivalent star-connected 
windings to give the same line currents and voltages. Then the 
line is designed and its performance calculated the same as in the 
preceding article. As a matter of fact, for line calculations it is 
only necessary to know the power, the voltage, and the power- 
factor of the load. The fact that the generator or the load is 
delta- or Y-connected has no bearing upon the line performance 
with a balanced load. 

Prob. 1. A 2000-kw. 6600-volt induction motor is fed from a 66,000- 
volt three-phase line through three step-down transformers, the high- 
tension windings of which are connected in Y, the low-tension windings 
in delta. What are the currents in these windings when the motor is 
carrying a 25 per cent overload? It is estimated that at this overload 
the power-factor is 90 per cent and the efficiency 92 per cent. The mag- 
netizing current of the transformers is negligible. 

Ans. 26.4 and 153 amp. 

Prob. 2. Show that, while the instantaneous electrical output of a 
single-phase alternator varies at double the frequency of the current, 
the output of a polyphase machine is practically constant as long as the 
load remains constant. Show that the same is true for motors. 

Note: For the electrical relations in two- and three-phase systems 
with unbalanced loads, and also for the theory of the V and T connections, 
see the author's Experimental Electrical Engineering, Vol. 2, Chapter 25. 
A more exhaustive treatment will also be found in his investigation 
entitled Ueber mehrphasige Stromsysteme bei ungleichmassiger Belastung 
(published by Enke, 1900). See also the chapters on polyphase systems 
in Dr. Steinmetz's Alternating-current Phenomena. 



CHAPTER XI 
VOLTAGE REGULATION OF THE TRANSFORMER 

38- Imperfections in a Transformer Replaced by Equivalent 
Resistances and Reactances. The reader is familiar in general 
with the construction and operation of the constant-potential 
transformer (Fig. 39). It consists of an iron core upon which 
two windings are placed as closely as possible to each other. 
When one winding is connected to a constant-potential alternat- 
ing-current source of power, an alternating magnetic flux is excited 
in the iron core and an alternating voltage is induced in the other 
winding. If this latter winding is connected to an electrical load, 
an alternating current flows through it, and causes a correspond- 
ing flow of current through the first winding, in order that power 
may be transmitted from the primary into the secondary circuit. 

The constant-potential transformer is one of the most perfect 
pieces of electrical apparatus, in that its efficiency (in medium and 
large sizes) is nearly one hundred per cent, and its voltage regula- 
tion with varying load is quite close. On the other hand, the 
requirements for voltage regulation are quite exacting, there being 
no provision in the apparatus itself for adjusting the voltage, like 
the field rheostat in a generator. Therefore, the pre-determina- 
tion of the voltage regulation of a transformer is of considerable 
practical importance. 

Numerically, the regulation of a transformer is expressed in 
a manner similar to that given in Art. 33 above for the trans- 
mission line. Let, for example, the rated secondary voltage of a 
ten-to-one transformer be 220 volts, and let us suppose that a 
primary e.m.f. of 2280 volts is necessary in order to have the 
rated secondary voltage at the rated load. Let now the second- 
ary circuit be opened; the secondary voltage will rise to prac- 
tically 228 volts, provided that the primary voltage is kept con- 
stant. Then, by definition, the regulation of the transformer at 
this load is 100 (228 - 220)/220 = 3.64 per cent. Let, in general, 

108 



Chap. XI] REGULATION OF THE TRANSFORMER 109 

the secondary terminal voltage at a certain load be E 2 , that at no 
load Eq2. Then, by definition, 

per cent regulation = 100 (E 02 - E 2 )/E 2 . . . (179) 

The difference between the no-load voltage and that at full load 
is due to slight imperfections in the transformer itself. There- 
fore, in order to be able to calculate the voltage regulation at a 
given load, it is necessary to learn the nature of these imperfec- 
tions; for purposes of computation, it is convenient to replace 
these imperfections by certain resistances and reactances, as shown 
in Fig. 39. 

In an ideal transformer the ratio of the primary to the second- 
ary voltage is equal to the ratio of the numbers of turns in the 
corresponding windings. The same relation is very nearly true in 
any good transformer at no load. This follows from the fact that 
the two windings are linked with the same magnetic flux, and 
hence the voltage induced per turn is the same in both. Hence, 
denoting the primary and secondary induced voltages by En and 
Ei2, and the corresponding numbers of turns in series by n\ and n 2 , 
we have 

Ea/E i2 = n x /n 2 (180) 

Furthermore, in an ideal transformer 

Ji*i = hn 2 , ........ (181) 

that is, the currents are inversely as the numbers of turns, or the 
primary ampere-turns are equal and opposite to the secondary 
ampere-turns. This is because an ideal transformer is supposed 
to have no reluctance in its magnetic circuit, so that no ampere- 
turns are required to maintain a magnetic flux in it. Consequently, 
any secondary current required by the load automatically draws 
a compensating primary current from the source of power, of such 
value that eq. (181) is satisfied. In a real transformer the primary 
ampere-turns are slightly different from the secondary ampere- 
turns, and the difference between the two is just sufficient to main- 
tain the alternating flux through the reluctance of the core, and to 
supply the core loss. Multiplying eqs. (180) and (181) term by 
term, and canceling n\ and n 2 , we find that 

Enli = Ei 2 I 2 . 

This simply means that an ideal transformer transmits power 
from the primary into the secondary circuit without loss. 



110 THE ELECTRIC CIRCUIT [Art. 38 

(a) The Ohmic Drop. One of the causes of the internal volt- 
age drop in a transformer is the ohmic resistance of its windings. 
Because of the resistance of the primary winding, the primary 
terminal voltage E x (Fig. 39) is slightly larger than the induced 




x x r x lA 



I^eE, S'ol Y 16 



B N 




■A/ww v - / oootfo<Ry 

r„ 



^2 



/ ~ 

Iron Core Windings 

Fig. 39. Imperfections in a transformer represented by resistances and 

reactances. 

counter-e.m.f. En which balances it. The secondary resistance 
causes a voltage drop, so that the secondary terminal voltage E 2 
is smaller than the secondary induced e.m.f. E i2 . Thus, the 
effect of the internal resistances upon the terminal voltages is 
such as to make the ratio E 2 /Ei smaller than the ratio n 2 /rti. 
The windings themselves may be thought of as devoid of resist- 
ance, but corresponding resistances n and r 2 may be placed out- 
side the transformer, as shown in Fig. 39. l 

(b) The Reactive Drop. Another imperfection or cause of 
internal voltage drop is the so-called leakage reactance of the 
windings. The total magnetic flux in a loaded transformer may 
be considered as consisting of three components; viz., the useful 
flux linking with both the primary and the secondary windings, 
the primary leakage flux linking with the primary winding only, 
and the secondary leakage flux linked with the secondary winding 
only. In an ideal transformer the two last-named fluxes are 
absent because the two windings are supposed to be perfectly 
interwoven, so as to leave no room for the leakage flux. The 
primary leakage flux, being produced by the primary current, is 
in phase with it, and induces an e.m.f. in lagging quadrature with 

1 The equivalent resistances ri and r 2 must replace not only the true 
ohmic resistances of the windings, but should also account for the eddy- 
current loss in the conductors. In low-tension windings made of heavy 
conductors, this latter loss may be at least as great as the theoretical i 2 r loss. 
In new transformers the eddy-current loss can only be estimated; in actu- 
ally built transformers it is calculated from the wattmeter reading on short 
circuit. 



Chap. XI] REGULATION OF THE TRANSFORMER 111 

this current. This e.m.f. must be balanced by part of the applied 
primary voltage, so that either this voltage or the induced e.m.f. 
En must be different from that in an ideal transformer. The 
effect of the secondary leakage reactance is similar, in that it 
absorbs part of the secondary induced voltage E i2 , and makes 
the secondary terminal voltage E 2 different from that in an 
ideal transformer. It is shown below that, with a load of lag- 
ging power-factor, the reactive drop in both windings lowers the 
secondary terminal voltage. With a leading secondary current, 
the reactive drop is in such a phase position as to raise the sec- 
ondary voltage. For purposes of computation, the transformer 
windings are assumed to produce no magnetic leakage fluxes, but 
imaginary reactance coils are connected in series with the wind- 
ings (Fig. 39). The reactances X\ and x 2 of these coils are such as 
to cause the same reactive voltage drop as that due to the actual 
leakage fluxes in the transformer. 1 

(c) The Exciting Admittance. Having thus made the windings 
of the transformer perfect by placing their impedances outside, 
we still have the problem of making the magnetic circuit ideal 
also. As stated before, the primary and secondary ampere-turns 
are not quite equal, because of a certain number of ampere-turns 
necessary to magnetize the iron. This means that a current 
must flow through the primary winding even when the secondary 
circuit is open. This current is called the no-load or magnetizing 
current of the transformer. Its amount depends upon the reluct- 
ance of the magnetic circuit and upon the core loss (hysteresis 
and eddy currents). For purposes of computation the iron core 
may be assumed to be of zero reluctance, and to have no core 
loss; but we may imagine a fictitious or equivalent susceptance 
b and a conductance g (Fig. 39) connected across the primary 
winding to draw a current equal in phase and magnitude to the 
exciting current of the transformer. Both g and b are shown 
connected across the induced voltage En, because both the mag- 
netizing current and the core loss depend upon the value of the 
flux and consequently upon the value of E ih which is proportional 
to the flux. Let the calculated or measured core loss be equal to 
Po watts; then g is determined from the equation 

Po = E il *g (182) 

1 For further details in regard to the leakage reactance of transformers, 
see the author's Magnetic Circuit, Art. G4. 



112 THE ELECTRIC CIRCUIT [Art. 39 

The pure magnetizing current, without the core-loss component, 
is in phase with the flux which it produces, and therefore is in 
quadrature with the induced voltage En. For this reason, it is 
represented as flowing through a pure susceptance. Knowing 
the pure magnetizing current /</, the susceptance b is determined 
from the equation 1 

b = I '/E il (183) 

Neither the core-loss component nor the pure magnetizing current 
are proportional to the flux or to the voltage E ih so that strictly 
speaking both b and g are functions of the counter-e.m.f. En. 
However, in practice, E a varies so little with the load that it is 
admissible to assume g and b Q to be constant quantities. More- 
over, the influence of the magnetizing current upon the voltage 
regulation is negligible in most cases. The magnetizing current 
is mentioned here only for the sake of completeness, so as to make 
the transformer core absolutely perfect. We shall see in the next 
chapter that b and g are of considerable importance in the per- 
formance of the induction motor. 

Thus, by the foregoing reasoning, both the core and the wind- 
ings of the transformer are made ideal, and all the imperfections 
are replaced by external resistances and reactances. Having done 
this, the performance of a transformer can be readily treated either 
graphically or analytically, as explained below. 

Prob. 1. Draw a diagram similar to Fig. 39 for a transformer with 
several secondary windings supplying independent load circuits. 

Prob. 2. Draw a diagram similar to Fig. 39 for an auto-transformer. 

39. The Vector Diagram of a Transformer. Having re- 
duced the transformer to an equivalent electric circuit (with a 
perfect magnetic link), the current and voltage relations at a 
certain load may be represented by a vector diagram (Fig. 40). 
In order to make the relations clearer, the voltage drop and the 
losses are greatly exaggerated. For this reason, the graphical 
treatment is more suitable for purposes of explanation than for 
numerical computations. For actual calculations the analytical 
method given in the next article is preferable. 

1 The calculation of the core loss and of the magnetizing current belongs 
properly to the theory of magnetic phenomena, and is treated in detail in the 
author's Magnetic Circuit, Arts. 19, 33, and 34. Here the values of P and 
/</ are supposed to be known. 



Chap. XI] REGULATION OF THE TRANSFORMER 



113 



Let the secondary terminal voltage and the load be given, so 
that the vectors E 2 and I 2 can be drawn in magnitude and rela- 
tive phase position. The secondary induced e.m.f. E i2 is found 




Fig. 40. The vector diagram of a transformer. 

by adding to E 2 the ohmic drop I 2 r 2 in phase with I 2 , and the 
reactive drop I 2 x 2 in leading quadrature with I 2 . 

The primary induced voltage En is in phase with E i2 , because 
both are induced by the same magnetic flux. The magnitudes 
of the two voltages are as the respective numbers of turns; see 



114 THE ELECTRIC CIRCUIT [Art. 39 

eq. (180). The vector marked En in Fig. 40 is in reality equal and 
opposite to En, and represents the part of the primary terminal 
voltage that balances En. Without the primary drop, the total 
applied primary voltage would be equal to En. But, on account 
of the primary drop in the transformer, the applied voltage E\ is 
obtained by adding to En the resistance drop I\T\ in phase with 
the primary current I\ and the reactive drop I\X\ in leading 
quadrature with I\. 

In order to be able to construct the vectors I\T\ and I\X\ it 
is necessary to know the vector of the primary current I\ in 
magnitude and phase position. In an ideal transformer, the 
primary current is in exact phase opposition to the secondary 
current, and the ratio of the two currents is inversely as the ratio 
of the respective numbers of turns; see eq. (181). In the actual 
transformer, the primary current, in addition to this component 
7 2 (n 2 /ni) " transmitted into the secondary," has a magnetizing 
component 7 , which serves to maintain the alternating flux in 
the core, and which is not transmitted into the secondary circuit. 
The total primary current is the geometric sum of the two com- 
ponents, and can be constructed if the magnetizing current I is 
known. 

The magnetizing current itself consists of two components, as 
explained in the preceding article, under (c). One component, 
Iq ', is in phase with the useful magnetic flux 0, and would be the 
only magnetizing component if the iron had no hysteresis and no 
eddy currents. This component is in quadrature with the induced 
voltage E ih and is, with respect to it, the reactive component of 
the magnetizing current. The other component, Iq" , in phase 
with En, represents a loss of power, and is therefore called the 
energy or loss component of the magnetizing current. Knowing 
Iq and Iq", the vector 7 is easily obtained. 

The vector of flux, 0, is drawn in leading quadrature with the 
induced e.m.f. E i2 , in accordance with Faraday's law of induction. 
If the flux varies according to the law 

t = <P m sm2wft, (184) 

the induced e.m.f. varies according to the law 

e i2 = —n 2 d4>t/dt = —2irfP m n2 cos 2 wft, . . (185) 
the second sine-wave lagging by 90 degrees behind the first. 



Chap. XI] REGULATION OF THE TRANSFORMER 115 

It is assumed in the construction of Fig. 40 that the primary and 
secondary inductive drops can be calculated separately. Such is, 
however, not the case with our present state of knowledge; both 
theory and experiment enable us to determine only the total re- 
active drop, including primary and secondary. Therefore, when 
it is desired to use the vector diagram for actual computations, it 
is customary to ascribe one half of the Ix drop to the primary and 
the other half to the secondary circuit. 

Usually, the magnetizing component of the primary current 
can be neglected; then it does not make any difference how the 
inductive drop is distributed. It will be shown in the next article 
that in such case the voltage regulation depends only upon the 
total impedance drop, either calculated or determined from a 
short-circuit test. When the internal voltage drop is given in 
per cent, it is understood to refer to the no-load voltage of each 
particular circuit. For instance^ if the reactive voltage drop in 
a 20/1-kv. transformer is said to be 5 per cent, this means that 
the secondary drop is 2.5 per cent of 1000 volts, or is equal to 25 
volts, and that the primary drop is 2.5 per cent of 20,000 volts, or 
is equal to 500 volts. 

Prob. 1. What is the regulation of a 600-kw., 2200/220- volt, 25-cycle 
transformer at the rated current and at 80 per cent power-factor (lag- 
ging)? The total reactive drop is 10 per cent, the primary ohmic drop 
is 2.2 per cent, and the secondary ohmic drop 2.8 per cent. The mag- 
netizing current may be neglected. 1 Ans. 10.1 per cent. 

Prob. 2. Determine the per cent voltage regulation of the trans- 
former specified in the preceding problem at the rated load and at 80 per 
cent power-factor, leading. 

Ans. — 1.4 per cent. The negative sign indicates a rise in secondary 
voltage, instead of a drop. 

Prob. 3. Correct the vector diagram of problem 1 for the magnetiz- 
ing current, knowing that the core loss amounts to 20 kw., and that 8500 
effective ampere-turns are necessary to maintain the flux, without the iron 
loss. The number of turns in the secondary winding is 64. 

Prob. 4. Adapt the diagram shown in Fig. 40 to an auto-transformer. 

40. Analytical Determination of Voltage Regulation. — Ap- 
proximate Solution. As explained above, it is preferable to 
calculate the voltage regulation of a transformer analytically, 

1 An excessive internal drop is selected purposely to enable the student 
to construct an accurate vector diagram to a convenient scale. The losses 
and the magnetizing current in problem 3 also are too high for a standard 
transformer. 



116 THE ELECTRIC CIRCUIT [Art. 40 

because the vectors of voltage drop are very small as compared 
with those of the primary and secondary voltages. The relations 
shown in Figs. 39 and 40 are expressed analytically by the 
equations 

E 2 = E i2 - I 2 Z 2 , (186) 

and 

Ei=Eu+!iZi (187) 

Since our purpose is to find the relation between E\ and E 2 , it 
is necessary to eliminate from these equations E it and E i2 . The 
relation between En and E i2 is given by eq. (180); therefore, we 
multiply eq. (186) by n\/n 2 and subtract it from eq. (187). The 
result is 

E x - (m/n 2 ) E 2 = ZjZi + (n 1 /n 2 )I 2 Z2. . . (188) 

The correct relation between Ii and I 2 is (Fig. 39) 

h = h (Wwi) + U = h + Jo, • • • (189) 
where 

lL = h(n2/nO (190) 

is the primary load current, or that part of the primary current 
which is transmitted into the secondary circuit. In a great 
majority of practical cases the magnetizing current is only a few 
per cent of the total primary current at the rated load. The 
voltage drop in the primary winding is also but a few per cent of 
the line voltage E\. For these reasons, it is permissible in Fig. 39 
to transfer the exciting admittance Y Q from the place MN to the 
primary terminals AB. The voltage drop in the transformer is 
then caused only by the load current, so that for the purpose of 
calculating regulation we may use the approximate relation 

I 1= I L = I 2 (n 2 / ni ) (191) 

Substituting for 1 1 and I 2 their values from eq. (191) in terms of. 
I l, we finally obtain 

E,-E L = 7l[Zx + (m/n 2 ) 2 Z 2 ] (192) 

In this equation, the quantity 

E L = {n l /n 2 )E 2 . . . . . . (193) 

is called the primary load voltage, or the secondary terminal volt- 
age reduced to the primary circuit. The expression (ni/n 2 ) 2 Z 2 is 
called the secondary impedance reduced to, or transferred into, 
the primary circuit. The quantity 

Z = Z x + (rn/ntfZ, (194) 



Chap. XI] REGULATION OF THE TRANSFORMER 117 

is called the total or equivalent impedance of the transformer 
reduced to the primary circuit. 

Using in eq. (192) the abbreviated notation introduced in 

eq. (194), we get 

E!-E L =I L Z (195) 

Equation (195) corresponds to the simplified equivalent diagram 
of the transformer shown in Fig. 41. This diagram differs from 
Fig. 39 in two respects: (1) The magnetic link is omitted, the 
primary circuit being connected directly to the modified second- 
ary circuit; (2) the exciting admittance is connected across the 
primary terminal voltage instead of across the induced voltage. 
The latter change makes the equivalent diagram only approxi- 
mately correct, but simplifies computations greatly. 

Equation (195) is identical in form with eq. (159), Art. 33, 
for the voltage drop in a transmission line; both are solved, and 
the per cent voltage drop determined, in the same way. In fact, 
without the exciting admittance F , the equivalent diagram shown 
in Fig. 41 reduces the performance of a transformer to that of a 
transmission line. 

Expression (194) for the equivalent impedance shows that 

resistances and reactances can be transferred from the secondary 

a x i r, u r.!, x i 



jr — z^TffiftHftffi^WVW — y 



1/ 

Line Ejgrf Y jfr, E n x=x x +x* E t Load 



-2 C 



B N D 

Fig. 41. The approximately equivalent diagram of a transformer or an 
induction motor. 

circuit into the primary, and vice versa, by multiplying them by 
the square of the ratio of the numbers of turns. For instance, in 
a 10,000/ 1000-volt transformer, a 1-ohm resistance in the low- 
tension circuit causes the same per cent voltage drop as a 100- 
ohm resistance in the high-tension circuit. This is easily verified 
as follows: Let the current in the low-tension circuit be 20 amp.; 
then in the high-tension circuit the current will be 2 amp. The 
drop in the 1-ohm resistance is 20 volts, or 2 per cent of the 
secondary voltage. The drop in the 100-ohm resistance is 200 
volts, which is 2 per cent of the primary voltage. In other words, 
the same reduction in the load voltage will be produced by using 



118 THE ELECTRIC CIRCUIT [Abt. 40 

either a resistance of one ohm in the secondary circuit or 100 ohms 
in the primary circuit. 

The secondary resistance r 2 transferred into the primary cir- 
cuit is denoted in Fig. 41 by r 2 , where 

rj = r 2 (ni/n 2 ) 2 (196) 

Correspondingly 

xi = X2 (n!/n 2 ) 2 (196a) 

The equivalent impedance Z consists of the quadrature sum 
of the equivalent resistance 

t'= n + rj = n + (n 1 /n 2 ) 2 r 2) .... (197) 

and the equivalent reactance 

x — xi + x 2 = xi + (ni/n 2 ) 2 x 2 .... (197a) 

The equivalent resistance is easily calculated, knowing the re- 
sistances of the two windings and the voltage ratio of the trans- 
former. Or else it is calculated directly from the i 2 r loss measured 
by a wattmeter in a short circuit test. The equivalent reactance 
is calculated from the terminal voltage in the short-circuit test, 
making a proper allowance for the known resistance drop. To 
illustrate, when the secondary circuit is short-circuited, eq. (195) 
becomes 

E 1 = I L Z (198) 

Ei and II are measured directly, so that Z can be calculated. 
Knowing the equivalent resistance r = P/Il 2 , the reactance is 
calculated from the expression x = Vz 2 — r 2 . For a new trans- 
former, the total equivalent leakage inductance is estimated with 
sufficient accuracy by means of various semi-empirical formulae; l 
or else the total impedance drop IlZ is taken as a certain per- 
centage of the rated voltage, from previous experience with similar 
transformers. 

Prob. 1. Check analytically the answers to problems 1 and 2 in the 
preceding article. 

Prob. 2. The high-tension winding of a 2000-kva., 33/1 1-kv. trans- 
former was short-circuited, and the voltage on the low-tension side ad- 
justed so as to circulate the rated current through the windings. The 
instrument readings were 470 volts and 30 kw. Calculate the per cent 
ohmic and reactive drops in the transformer. Ans. 1.5 and 4 per cent. 

Prob. 3. Deduce a formula similar to (192), but referring to the 
secondary circuit. 

1 See for instance the author's Magnetic Circuit, Art. 64. 



Chap. XI] REGULATION OF THE TRANSFORMER 119 

Prob. 4. Show how the voltage regulation of a transformer can be 
estimated, using Mershon's diagram given in various electrical handbooks 
and pocketbooks. 

Prob. 5. The primary voltage of a given transformer is kept constant 
at a known value. Determine the percentage internal drop (Ei — El) I E t 
for a given impedance of the load. 1 Solution: Let the load impedance, 
reduced to the primary circuit, be Zl) then the load current is 

iL^Ei/fZL + Z), 

where Z is the equivalent impedance of the transformer itself, supposed 
to be known. The load voltage is 

E L = E X - ZI l = E 1 Z L /{Z L + Z) = E 1 /[l + (Z/Z L )]. 

Having expressed all the known and unknown quantities in the complex 
form, in either Cartesian or polar coordinates, the magnitude and direc- 
tion of El can be determined, by using the general method, i.e., equating 
the real and the imaginary parts on both sides of the equation. 

Prob. 6. The equation for El given in the preceding problem leads 
to involved numerical computations. Moreover, the difference Ei — El 
cannot be accurately determined in this way when El differs but little 
from Ei. Show how to simplify the numerical work, by taking advan- 
tage of the fact that Z is small compared with Zl. Solution : When a 
quantity a is small compared to unity, we have by division 1/(1 -f- a) = 
1 — a + a 2 — etc. We have accordingly 

E L = E 1 [l- (Z/Z L )\ approximately, 
or 

E L = E 1 -E 1 Z/Z L (A) 

Let El be the vector of reference; consequently E± = Ei (cos + j sin 0). 
Let also Zl = zl (cos 4>l + j sin <f>L) and Z = z (cos </> + j sin 4>). Then 
according to eqs. (154) and (156), 

E X Z!Z L = Eyz/ZL [cos (0 + </> - 4>l) + j sin (d + <t> - <f> L )], 

or, denoting E x Z/Zl by Ai?i and 4> — 4>l by 13, we have 

aEi = AEi [cos (0 + 0) + j sin (0 + £)]. 

Equation (A) may now be written in the form 

E x (cos + j sin d) = E L + aEi [cos (9 + 0)+ j sin (0 + 0)], 
where 

A#i = Etf/ZL 

is a known quantity, as well as the angle = <f> — <i>L. Separating the 
real and the imaginary parts, we get 

#icos0 = E L + aE 1 cos(0 + 13), (B) 

Ei sin = AEi sin (d + p) = AE X sin cos /s + A#i cos sin p. (C) 

1 The conditions in this problem differ from those in the text above in 
two respects: (1) The primary voltage is given instead of the secondary; 
(2) the load is given by its impedance instead of the current and power- 
factor. 



120 THE ELECTRIC CIRCUIT [Art. 41 

From eq. (C), dividing both sides by cos 6, we find 

tan0 = A#isin/3/(i?i — AZ?iCOS/3), . . . . (D) 

from which d can be calculated. Using in eq. (B) the transformation 
cos = 1 — 2 sin 2 \ 0, the same as in Art. 33, we get, after division by E h 

(Ei - E L )/Ei = (AEi/E!) cos (9 + 0) + 2 sin 2 } d. . . (E) 

While the derivation of formulae (D) and (E) may seem somewhat tedi- 
ous, the results are in the form most convenient for numerical work. 

41. Analytical Determination of Voltage Regulation. — Exact 
Solution. 1 The approximation made in the preceding article 
consists in shifting the exciting admittance F so that it is con- 
nected across the primary terminal voltage E h instead of across 
the primary induced voltage En (compare Figs. 39 and 41). 
Retaining the exciting admittance in its correct place, we obtain 
the equivalent diagram shown in Fig. 42. The secondary im- 

A x i r i M r * x 'i r> 

- ^5 -h 



Line 



Ii 



W Y |5 E Load 



B N D 

Fig. 42. The correct equivalent diagram of a transformer or an 
induction motor. 

pedance is reduced to the primary circuit as before, by being 
multiplied by the square of the ratio of turns (ni/n 2 ) 2 . This pro- 
cedure is strictly correct, the magnetic link being by assumption 
perfect. 

Equations (186) and (187) hold here as before, but instead 
of using the approximate eq. (191) we shall use the correct rela- 
tion (189). The magnetizing current is 

-Jo^E-aYo, . (199) 

so that the total primary current is 

U= !l+EhYo (200) 

Expressing I\ and I 2 in eqs. (186) and (187) through the load 
current I l, and eliminating E ix and E i2 as before, we obtain 

(E, - Zil^/a + ZiYo) =E L + I L (m/ntfZ*. (201) 

This equation takes the place of the approximate eq. (192). The 
two equations become identical when Y = 0. 

1 This article may be omitted if desired. 



Chaq. XI] REGULATION OF THE TRANSFORMER 121 

The complex quantity 1 -f ZiY which enters into eq. (201) 
may be called a correction factor, and may be represented in the 
form 

K = 1 + Z 1 Y = k (cos a + j sin a). . . . (202) 

Since the ratio of two complex quantities is also a complex quan- 
tity, eq. (201) may be expressed in the form 

E cl =E L +I L Z c , (203) 

where the corrected primary voltage 

E cl = E./K = (Et/k) [cos (6 - a) + j sin (0 - a)], (204) 
and the corrected equivalent impedance 

Z c = Zx/K + (ni/n 2 ) 2 Z 2 = z c (cos yp + j sin ^). . (205) 

Equation (203) is of the same standard form as eqs. (195) and 
(159), and can be solved by the method given in Art. 33. 

This problem can be solved also by keeping the complex quan- 
tities in the orthogonal form. The student will profit by working 
out the details for himself. 

Sometimes it is desired to know the voltage regulation of a trans- 
former over a certain range of loads, the results being represented 
in the form of a curve. In such a case, it makes no difference 
for which particular loads the regulation is actually calculated, 
provided that these loads are selected within certain limits. 
If the primary voltage is given and is constant, it may be more 
convenient to perform the calculations (according to Fig. 42), 
not for an assumed current 1 2 , but for an assumed load imped- 
ance Zl. Combining the impedances in series and the admit- 
tances in parallel, the whole circuit connected at the primary 
terminals is finally reduced to one impedance. Dividing the 
primary voltage by this impedance gives the primary current, and 
consequently the drop in the primary impedance Z\. Thus, the 
voltage En becomes known, and the current I Q can be calculated. 
After this, the current I l and the drop I lZ 2 are calculated. This 
drop, being subtracted from En, gives the desired secondary 
voltage E 2 , reduced to the primary circuit. 



CHAPTER XII 

PERFORMANCE CHARACTERISTICS OF THE 
INDUCTION MOTOR 

42. The Equivalent Electrical Diagram of an Induction Mo- 
tor. The student is supposed to be familiar with the general (quali- 
tative) explanation of the performance of a polyphase induction 
motor, and with the general shape of the load characteristics. 1 
It will be shown here how to predetermine the performance char- 
acteristics of a given induction motor by reducing it to an equiva- 
lent electric circuit, similar to that of a transformer. 

The following experiment shows the possibility of such an 
equivalent diagram. A brake test is performed on the motor, 
and the primary current and the power-factor are plotted against 
the output as abscissae. Then the rotor is blocked, and variable 
non-inductive resistances are inserted into its phase windings. 
If the rotor has a squirrel-cage secondary, resistances must be 
inserted in series with each bar, or into each section of the end- 
rings between consecutive bars. The motor is thus reduced to a 
polyphase transformer, the inserted secondary resistances repre- 
senting the load. A load test is performed on this transformer, 
and the curves of primary current and power-factor are plotted 
against the total i 2 r loss in the external resistances. These curves 
are found to coincide very closely with the curves obtained from 
the brake test, provided that the brake power and the i 2 r power 
are plotted to the same scale, one representing the mechanical, 
the other the corresponding electrical output. Some difference 
in the curves is due to the fact that the stationary transformer 
has no friction loss; this is, however, partly or wholly compen- 
sated by a greatly increased secondary core loss. 

The theoretical reasons for this equivalence of an induction 
motor to a polyphase transformer will become clear by consider- 

1 See, for instance, the author's Experimental Electrical Engineering, Vol. 1, 
chap. 17. 

122 



Chap. XII] THE INDUCTION MOTOR 123 

ing the electrical relations in the rotor under the following three 
headings : 

(a) The Relationship between the External Resistance and the 
Slip. Let the input into the rotor be P watts per phase of the 
secondary winding, and let the motor be running at a slip s. For 
instance, s = 0.05 means that the speed of the rotor is 5 per cent 
lower than the synchronous speed, or the speed of the revolving 
field. Then, sP watts are converted into heat in each phase of 
the secondary winding, and (1 — s)P watts are available on the 
shaft as the output (including friction and windage). This is 
because the tangential electromagnetic effort is the same on the 
surface of the stator as it is on the surface of the rotor. But 
while the gliding magnetic flux travels at synchronous speed, the 
rotor travels at (1 — s) times the synchronous speed. The elec- 
tromagnetic coupling between the stator and the rotor is simi- 
lar to a friction coupling between two shafts, having a certain 
amount of slip. If the speed of the driven shaft is say 5 per cent 
below that of the driving shaft, on account of the slip in the 
coupling, 95 per cent of the power is transmitted and 5 per cent 
is lost in heat in the coupling. 

With the rotor blocked, let R be the external resistance per 
phase of the secondary, and let I 2 be the secondary current per 
phase. For a slip s we must have the condition 

s (R + r 2 ) J 2 2 = r 2 I 2 \ 
or 

s = r 2 /(R + r 2 ) (206) 

If the slip is given, the required external resistance is 

R = r 2 (l - s)/s . (207) 

(b) Equal Secondary Current and Phase Displacement with the 
Rotor Running or Blocked. Let the reactance of the secondary 
winding per phase be x 2 ohms, at the primary or synchronous 
frequency. With the rotor running at a slip s, the frequency of 
the secondary currents is only equal to s times the primary fre- 
quency, so that the reactance per phase is sx 2 . Therefore, the 
phase displacement </> 2 between the induced secondary voltage 
and the current is determined by the relation 

tan </> 2 = sx 2 /r 2 (208) 

With the rotor blocked and provided with external resistances 
satisfying condition (206), the total resistance of the secondary 



124 THE ELECTRIC CIRCUIT [Art. 42 

circuit per phase is R + r 2 = r 2 /s. The secondary frequency 
is equal to that in the primary circuit, so that 

tan 02 = x 2 /(r 2 /s) = sx 2 /r 2 ; 

thus the phase displacement is the same as that given by eq. (208). 
With the same revolving magnetic flux in both cases, the currents 
are also equal. While with the stationary rotor the induced 
secondary e.m.f . is larger in the ratio of 1 : s, because of a higher 
speed of cutting the secondary conductors, yet the total secondary 
resistance R + r 2 is also larger in the same ratio of 1 : s, according 
to eq. (206). The secondary reactance is also larger in the same 
ratio on account of the higher frequency. Thus, with the rotor 
blocked, both the e.m.f. and the impedance of the secondary 
circuit are larger in the ratio of 1 : s than when it is running at a 
slip s. Hence, the current, which is equal to the ratio of the e.m.f. 
to the impedance, is the same in both cases. 

(c) The Reaction of the Secondary upon the Primary Circuit 
is the Same with the Rotor Running or Blocked. The magneto- 
motive force of the revolving rotor is the same as that of the 
stationary rotor with the resistance R in series, provided that in 
both cases the magnetomotive force is considered with respect to 
the stationary primary circuit. In the latter case the frequency 
of the secondary currents is equal to that of the supply, so that 
the resultant magnetomotive force due to all the secondary phases 
travels in the air-gap at synchronous speed, the same as the 
resultant magnetomotive force of the primary currents. The 
two magnetomotive forces form one resultant magnetomotive 
force which produces the revolving flux. With the revolving 
rotor, the frequency of the secondary currents is s per cent of that 
of the supply, so that the secondary magnetomotive force glides 
relatively to the body of the rotor at a speed equal to s per cent 
of the synchronous speed. But the speed of the rotor itself is 
the (1 — s) part of the synchronous speed. Hence, the velocity 
of the secondary magnetomotive force with respect to the stator is 
s + (1 — s) = 1, or is equal to the synchronous speed, and is 
the same as the velocity of the primary magnetomotive force. 
We have seen above that the secondary currents and their phase 
relation are the same in the two cases, so that the secondary 
magnetomotive force is also the same in phase and magnitude. 
Consequently, with the same flux, determined by the applied volt- 



Chap. XII] THE INDUCTION MOTOR 125 

age, the primary magnetomotive force is also the same in both 
cases. This means that the primary current and power-factor 
are the same with the stationary rotor loaded electrically as with 
the revolving rotor loaded mechanically. 

We have thus proved theoretically, as well as experimentally, 
that the performance of an induction motor may be reduced to 
that of a stationary transformer. But we know from the pre- 
ceding chapter that a transformer can in turn be replaced by an 
equivalent electric circuit, either approximately (Fig. 41) or accu- 
rately (Fig. 42). Thus, the same equivalent diagrams can be used 
in the predetermination of the performance of an induction motor. 
All the quantities which enter into these diagrams are understood 
to be per phase of the primary circuit (usually per phase of Y 
in a three-phase motor). When the number of the secondary 
phases and the method of connections are different from those 
in the primary circuit, the secondary winding is replaced by an 
equivalent one of the same number of phases, and with the same 
kind of connections as in the primary circuit; see Art. 45 below. 1 

Prob. 1. Explain the principle of the speed control of an induction 
motor by means of adjustable external resistances in the secondary cir- 
cuit. 

Prob. 2. Explain the principle of the direct and differential cascade 
connection of two induction motors. 

Prob. 3. Show how an induction generator can be reduced to an 
equivalent electric circuit. 

43. The Analytical Determination of Performance. — Ap- 
proximate Solution. The problem is to calculate the perform- 
ance characteristics of a given induction motor — in other words, 
against the output as abscissae, to plot the following curves; viz., 
primary amperes, kilowatts input, primary power-factor, slip, 
torque, and efficiency. The resistances and the leakage reactances 
of both windings, reduced to the primary circuit, are supposed to 
be known, so that each primary phase of the motor can be replaced 
by either the approximate or the exact diagram (Figs. 41 and 42). 
The approximate diagram only is considered here, because it is 
sufficiently accurate for most practical purposes. The exact solu- 
tion is given in Art. 47 below. The iron loss, friction, and the 

1 The values of the leakage reactances of the windings are supposed here 
to be known; for their calculation from the dimensions of the motor see the 
author's Magnetic Circuit, Art. 66. 



126 THE ELECTRIC CIRCUIT [Art. 43 

magnetizing current are also supposed to be known, so that the 
exciting admittance F is known. While in reality it varies some- 
what with the load, in the approximate solution it is considered 
to be a constant quantity. 

The problem is solved similarly to that of the voltage regula- 
tion of a transmission line or of a transformer, treated above; 
that is, a load current II is selected, and the circuit is solved in 
the complex notation. In order to make the treatment independ- 
ent of the other chapters, a complete solution is given below, with 
some minor changes which simplify the numerical work. As in 
the transmission line and in the transformer, we have 

Ei=E L + ZI L , (209) 

or, expanded, 

Ei (cos 6 + j sin 6) = E L + II (r + jx) . . . (210) 
Here the direction of the unknown load voltage El is again selected 
as the reference axis. The current II is in phase with El, because 
by assumption the external resistance R is non-inductive. Sepa- 
rating the real and the imaginary parts, we get 

EiCosd = E L + I L r; (211) 

Ei sin = I L x. . . . .... (212) 

When plotting the curves, it is immaterial which values of the load 
are selected for computation. We assume, therefore, a series of 
reasonable values for II, and from eq. (212) calculate the corre- 
sponding values of sin 6. Then from eq. (211) we find the values 
of El, and finally determine the outputs per phase from the 
equation 

Pl = IlE l (213) 

Knowing II, El, and the angle 0, the rest of the values for the 
performance curves are calculated as follows: 

(a) The Slip. The external resistance, reduced to the primary 
circuit, is 

R' = E L /I L ; • • (214) 

and the slip is found from the equation 

s = ! yj/{R' + r 2 '), ... . . . (214a) 

which is identical with eq. (206) , except that r 2 ' and R' are second- 
ary quantities reduced to the primary circuit. 

(b) The Primary Current and Power-factor. The total primary 
current per phase is 

!i=!o+!l, (215) 



Chap. XII] THE INDUCTION MOTOR 127 

where the magnetizing c-urrent J is known, and can be repre- 
sented with respect to the terminal voltage E x as 

I = I (cos 0o — jsin0 o ) (216) 

The load current, in its phase relation with respect to the terminal 
voltage, is 

!l = II (cos 6 - j sin 0), (217) 

so that 

Ii = ii—jii = (iz, cos 0+ Jo cos (/>o)— jCTlsui 0+/ o sin0 o ). (218) 
Knowing the projections i\ and i\ of the 'primary current with 
respect to the terminal voltage E\, the primary phase angle 0i 
is found from the equation 

tan </>i = ii/i h (219) 

and then the primary power-factor, cos <f> 1} is taken from a trigo- 
nometric table. The current itself, 

h = ii/cos 0i (220) 

(c) The input per phase is 

Pi = EJi cos 0i = Ei%! (221) 

The efficiency is equal to the ratio of the output to the input. 

(d) The useful torque in synchronous watts, or the input into 
the secondary, is equal to the output plus the secondary copper 
loss. The tangential effort per phase, in kilograms at a radius of 
one meter, or the torque per phase, in kg.-meters is 

T = 973.8 (P L + 0.001 iz,V)/(synchr. r.p.m.), . (222) 
Pl being expressed in kilowatts, so as to avoid large numbers in 
numerical applications. 

Sometimes the performance data are desired for one particular 
load, Pl, only; for instance, at the rated output of the machine. 
The method outlined above may in this case be somewhat tedious, 
because one has to find by trials the proper values of II and El 
which give the desired output. It may lead more quickly to the 
desired end to solve eqs. (211), (212), and (213) as three simul- 
taneous equations for the unknown quantities El, II, and 6. 
Squaring the first two equations and adding them together, the 
angle 6 is eliminated, and we get 

E 1 z = E L 2 + lL 2 z 2 + 2E L lLr. 
Substituting for El its value from eq. (213), gives a quadratic 
equation for zIl 2 , namely 

(zI L 2 ) 2 -2 {zIl 2 ) ih ES - P L r)/z~+ P L 2 = 0. . . (223) 



128 THE ELECTRIC CIRCUIT [Art. 43 

The solution of this equation is 

zI L 2 = (!#! 2 - P L r)/z - VQEi*-Pir)*/*-PL*- 

The minus sign only is retained before the radical, because it 
gives a smaller current. It can be shown that the solution with 
the plus sign corresponds to the unstable region of operation of 
the motor. 

For numerical computations the preceding equation is put in 
the form 

zI L 2 = 1000 (Q - Vq* _ p L 2 }j _ m m (224) 

where, for the sake of brevity, we introduce the notation, 

Q = (500 ^i 2 - P L r)/z. ..... (225) 

In the last two equations Pl and Q are in kilowatts, so as to avoid 
large numbers, and E\ is in kilovolts. The student is reminded 
that Ei is the phase or star voltage, and not the line voltage, and 
that Pl is the output per phase. 

When Pl is small compared to Q, formula (224) represents 
the difference of two quantities of nearly equal value. The 
result is inaccurate, and it is better to expand the expression 
[1 — (Pl/Q) 2 ]*> according to the binomial theorem. This gives 

zli 2 = 1000 Q ft (Pl/Q) 2 + I (Pl/QY + tV (Pl/QT + etc.]. (226) 

The latter formula is much more convenient for numerical applica- 
tions than eq. (224), because the second term in the brackets is 
small as compared to the first, and the third term can usually be 
neglected. 

Knowing II, the rest of the values are determined as before. 

Prob. 1. Plot complete performance curves of a three-phase, 25- 
cycle, 150-kw., 6-pole, 2200-volt, induction motor between no load and 
25 per cent overload, from the following data: Total no-load input (for 
all three phases) is 10.5 kw.; the no-load current per phase of the line 
is 13 amp. With the armature blocked, the input is 230 kw., the cur- 
rent per phase being 227 amp. 1 The resistance of the primary winding 
per phase of Y is 0.69 ohm. Hint: Follow consistently the approximate 
diagram, Fig. 41; that is, do not correct the no-load reading for the 
primary i 2 r loss, and assume the magnetizing current with the armature 
locked to be the same as at no load. 

1 The data with the armature locked refer to the rated voltage; they 
are obtained by extrapolating the curves taken at lower voltages. It would 
not be practicable to apply the full line voltage to a large motor with the 
armature blocked. 



Chap. XII] THE INDUCTION MOTOR 129 

Ans. At the rated output the primary current is 49 amp.; the 
power-factor is 90.8 per cent; the slip, 3.5 per cent; the efficiency, 88.5 
per cent; and the torque, 302 kg.-m. 

Prob. 2. Check the answer to the preceding problem, using eq. (226). 

Prob. 3. Extend the theory and the formulae given above to the 
performance characteristics of an induction generator. 

44. Starting Torque, Pull-out Torque, and Maximum Out- 
put. When judging the performance of a given induction 
motor, or designing a new motor, the following features are of 
importance : 

(a) The starting torque, either in its absolute value, or in 
its ratio to the torque at the rated load. If the motor is to be 
started by means of resistances in the secondary circuit, one may 
be required to calculate the values of these resistances necessary 
for a prescribed starting torque, or for a maximum starting 
torque. 

(b) The pull-out torque, or the torque at which the motor 
reaches the limit of stable operation, and comes to a stop. This 
torque is usually given through its ratio to the full-load torque. 

(c) The maximum output of the motor, in kilowatts. This 
output takes place at a smaller slip than that at which the motor 
pulls out. The output is a maximum when the product of 
torque times speed is a maximum, but not when the torque 
itself is greatest. 

The three' quantities mentioned above can be determined by 
using the equations deduced in the foregoing article. 

(a) The Starting Torque. In the general formula (222) , Pl = 
at start, because the motor supplies no mechanical output, the 
speed being equal to zero. In the equivalent electrical diagram 
(Fig. 41) this corresponds to a short curcuit of the load, or R = 0. 
Hence, II = Ei/z; substituting this value into eq. (222), we find 
that the starting torque per phase, in kg.- m. 

T st = 0.9738 EiW/iz 2 X synchr. r.p.m.). (227) 

It will be seen from this expression that the starting torque is 
proportional to the square of the line voltage. This fact permits 
one to determine the initial torque when starting a motor on a 
lower voltage, by means of auto-transformers. The same equa- 
tion shows that the starting torque increases with the secondary 
resistance. It is not quite proportional to it, because r 2 r is also 
implicitly contained in z. 



130 THE ELECTRIC CIRCUIT [Art. 44 

When the motor has a phase-wound secondary and is started 
by means of resistances in the secondary circuit, r 2 in formula 
(227) includes this secondary resistance. It is thus possible to 
calculate the external resistance required for a given starting 
torque. For example, when the starting torque is given, the 
ratio z 2 /r 2 in eq. (227) is a known quantity, or 

z 2 /r 2 ' = [(ri + r 2 y + x 2 ]/r 2 ' = c, . . . (228) 
where 

c = 0.9738 E x 2 / (desired torque X synchr. r.p.m.). (229) 

Solving the quadratic (228) for r 2 , we obtain 

ra ' = (J c - r'i) ± V^c-n) 2 - (x 2 + n 2 ). . (230) 

In applications, the minus sign only is retained before the radical 
because one would naturally use the smaller of the two resistances 
which give the same torque. 

The value of r 2 determined from this expression comprises 
both the resistance per phase of the rotor proper and the starting 
resistance per phase, if any is used, both reduced to equivalent 
primary values. To obtain their actual values, see Art. 45 below. 

If the external resistance is to be so selected as to give a max- 
imum starting torque, c in expression (228) must be a minimum. 
Equating to zero the derivative of c with respect to r 2 ', and 
solving for r 2 , we get 

rj = (x 2 +.ri»)*, (231) 

that is, r 2 is very nearly equal to x. This value comprises the 
resistance of the rotor proper and the starting resistance, both 
per phase of the primary circuit. If a resistance is selected which 
is either less than or greater than that determined by eq. (231), 
the motor does not develop its full starting torque. This checks 
with eq. (230), which shows that the same torque can be obtained 
with two different values of starting resistance. 

(b) Pull-out Torque. According to eq. (222), the torque is a 
maximum when 

E l Il + IlW = max (232) 

Here El and II are functions of the independent variable 0. 
Expressing them through 6 from eqs. (211) and (212), and omit- 
ting the constant factor £\ 2 , we obtain 

(1/x) sin [cos 6 — (r/x) sin 6] + (r 2 '/x 2 ) sin 2 6 = max., 



Chap. XII] THE INDUCTION MOTOR 131 

or, after simplification, 

zsin2 - r L (l - cos2 0) = max. . . . (233) 

Equating to zero the derivative of this expression with respect to 
0, gives 

x cos 2 — fi sin 2 = 0, 
or 

tan 2 = x/ri (234) 

Knowing 0, the values of El and II are calculated from eqs. (211) 
and (212), and then the torque is determined from eq. (222). 

It is of interest to note that the angle 0, at which the motor 
pulls out of step, is independent of the secondary resistance 
r 2 '. Neither does this resistance enter into eq. (233). Hence, 
the maximum torque which a motor is capable of developing is inde- 
pendent of its rotor resistance. This resistance determines only the 
speed at which the maximum torque takes place. The higher the 
secondary resistance, the lower the speed at which the motor 
pulls out of step. By using an external starting resistance, and 
a rotor winding of low resistance, two maxima of the torque are 
obtained, one at the start, with the external resistance in, and the 
other near synchronism, with it out. 

(c) Maximum Output. The problem is to find the values of 
El and II for which the product ElIl is a maximum. Again 
expressing El and II through the angle from eqs. (211) and (212), 
and omitting the constant factor Ei 2 /x, we have 

sin [cos — (r/x) sin 0] = max. 

Equating to zero the derivative of this expression with respect to 
0, gives 

e = H, (235) 

where tan <f> = x/r. Knowing the angle 0, the values of II and 
El are calculated from eqs. (211) and (212), and then their product 
ElIl is determined. 

Prob. 1. The motor specified in problem 1 of the preceding article 
is designed to be started at a reduced voltage. What per cent tap should 
be used on the auto-transformers in order to get a starting torque of 
about 30 per cent of the full-load torque? 

Ans. 60 per cent of the line voltage. 

Prob. 2. The same motor is provided with a phase-wound secondary 
and is to be started by using resistances in series with the rotor windings. 



132 THE ELECTRIC CIRCUIT [Art. 44 

What external resistance is necessary in order to obtain a starting torque 
equal to 1.5 times the full-load torque? 

Ans. 0.78 ohm per phase, in terms of the primary circuit. 

Prob. 3. What starting resistance in the preceding problem would 
give the maximum starting torque? Ans. About 5.7 ohms. 

Prob. 4. Show that the motor specified in problem 1 of the preceding 
article pulls out of step when the torque exceeds 2.45 times the rated 
full-load torque. 

Prob. 5. Check the answer to problem 4 by using the answer to prob- 
lem 3. 

Prob. 6. Show that the maximum output of the same motor is equal 
to 2.15 times the rated output 

Prob. 7. Show that the input into an induction motor is a maximum 
when = 45°. Hint : II cos 6 = max. 

Prob. 8. Show how to calculate the per cent slip at which the motor 
pulls out of step, and also the speed at which the output is a maximum. 



CHAPTER XIII 

PERFORMANCE CHARACTERISTICS OF THE 
INDUCTION MOTOR— (Continued) 

45. The Secondary Resistances and Reactances Reduced 
to the Primary Circuit. It is proved in Art. 40 that in a trans- 
former the secondary resistance and reactance can be transferred 
into the primary circuit by being multiplied by (ni/n 2 ) 2 . The 
same rule holds true for the induction motor, provided that the 
number of phases is the same in the primary and in the secondary 
windings, and that the two windings are of the same type (the 
same number of slots per phase and the same winding pitch). 
This is hardly ever the case, and with a different number of 
phases and different types of winding in the primary and second- 
ary, the following formula holds true : 

rV/r 2 = (mi/m 2 ) (k bl ni/k b2 n 2 ) 2 , .... (236) 
and analogously for the reactances, 

x 2 /x 2 = (wi/m 2 ) (hini/k b2 n 2 ) 2 . . . . (237) 

In these expressions, m stands for the number of phases, n is the 
number of turns per phase, and k b is the so-called breadth factor 
which characterizes the winding. The subscripts 1 and 2 refer 
to the primary and secondary windings respectively. The quan- 
tities r 2 and x 2 are the actual resistance and reactance per phase 
of the secondary circuit; r 2 and x 2 are the equivalent quantities 
per phase of the primary circuit. 1 When mi = m 2 and k bl = k b2 , 
the preceding formula? become identical with eqs. (196) and (196a) 
for the transformer. 

Equations (236) and (237) refer to the resistances and react- 
ances per phase, with the understanding that the windings of 
each phase are all in series, both in the stator and in the rotor; 
and that the connections are either both star, or both mesh, even 
if the number of phases be different. Otherwise, the actual con- 

1 For a proof of these formulae, see the author's Magnetic Circuit, Art. 44; 
the values of k b will be found in Arts. 27 to 29 of the same book. 

133 



134 



THE ELECTRIC CIRCUIT 



[Art. 45 



nections must be taken into consideration and the values of r 2 ' 
and X2 further modified, keeping in mind the fact that the total 
i 2 r loss must be the same in the equivalent winding as in the ac- 
tual one. As a simple illustration, let the stator be three- 
phase Y-connected, and the rotor three-phase delta-connected. If 
the equivalent secondary resistance calculated by means of 
eq. (236) is r 2 ', then only | r 2 ' must be used in the equivalent 
diagram, per phase of Y. This is because the current per phase 
of Y is V3 times as large as that per phase of delta, hence, for 
the same i 2 r loss, the resistance per phase of Y must be only one- 
third of that per phase of delta. The same relation holds true 
for reactances, because the stored electromagnetic energy is also 




e^>" 



Fig. 43. A squirrel-cage rotor and the Fig. 44. The vectorial relation between 
star resistances equivalent to the end the star and mesh currents in a sym- 
ring. metrical w-phase system. 

proportional to the square of the current (Art. 20), and the 
equivalent diagram must express correctly the cyclic exchange 
of energy between the primary and the secondary circuits. 

In the most general case, let an m-phase symmetrical system 
be given, for instance a two-pole squirrel-cage rotor (Fig. 43), 
and let it be required to find the relation between the mesh resist- 
ances r m and the star resistances r 8 such that the i 2 r loss per phase 
shall be the same in both. The relation between the vectors of 
the star currents and those of the mesh currents is shown in Fig. 44, 
the star currents I 8 forming an m-sided polygon, and the mesh 
currents I m being the radii of the polygon. This diagram is correct 
because it satisfies the following conditions: (a) the star currents 



Chap. XIII] THE INDUCTION MOTOR 135 

are displaced in phase relatively to each other by equal angles 
of 2r/m, over the whole range of 2t; (b) the same is true for 
the mesh currents; (c) each star current is the geometric differ- 
ence of the two adjacent mesh currents; (d) the geometric sum of 
the star currents is equal to zero (Kirchhoff's first law). From 
the geometry of the figure we have 

| /. = I m sin fr/m) (238) 

The condition I s 2 r s = I m 2 r m leads to the ratio 

Tm/fs = (7 8 /^) 2 = 4sin 2 (7r/m). . . . (239) 

A similar relation holds for the reactances. When m = 3, we 
find as before that r A /ry = 3. 

Equation (239) finds its practical application in the calcula- 
tion of the equivalent resistance and reactance of a squirrel-cage 
rotor. The sections of the end-rings between the bars are mesh- 
connected, while the bars themselves may be considered as parts 
of a star-connected ra 2 -phase system, where m 2 is the number of 
bars per pair of poles. Let n be the resistance of each bar, includ- 
ing the two contact resistances between the end-rings and the 
bar; let r r be the resistance of a section of an end-ring between 
two consecutive bars. The resistance of the rings can be replaced 
by added resistances in series with the bars, so as to change the 
connections to a pure star system (Fig. 43). According to eq. 
(239) we find that the new resistance per bar must be equal to 
r b + 2 77/ [4 sin 2 {ir/m^)]. If the motor has p poles, or J p pairs of 
poles, there are \ p bars in parallel belonging to the same phase, so 
that the total resistance of the secondary winding per phase is only 
the 2/p part of that of one bar. Hence, assuming that the primary 
winding is star-connected, and that all the coils in each phase are 
in series, the value of r 2 to be used in eq. (236) is 

r 2 = [r & + Jr r /sin 2 (7r/m 2 )]/(|p). . . . (240) 
Analogously, 

x 2 = [x b + i av/sin 2 (7r/m 2 )]/(i p). . . . (241) 

Since there is only one bar per phase, and one bar is equivalent 
to one-half of a turn, the value n 2 = | and k b2 = 1 must be 
used in eqs. (236) and (237). 

Prob. 1. A two-phase induction motor has the primary winding ar- 
ranged for two independent phases; the secondary is three-phase Y-con- 



136 



THE ELECTRIC CIRCUIT 



[Art. 46 



nected. When the rotor is stationary and its circuits are open, 440 volts 
impressed at the primary terminals produce 97 volts between the slip- 
rings. The calculated starting resistance, per phase of the primary 
circuit, is 14 ohms. What is the actual resistance to be used in series with 
the rotor windings? Hint : Consider the primary circuit as a four-phase 
star-connected system, so that nikbi/n 2 kb2 = \ X 440/(97/ V3). 

Ans. 0.34 ohm. 
Prob. 2. A six-pole, three-phase, Y-connected induction motor has a 
squirrel-cage rotor of 80 cm. diameter with 73 bars; the resistance of 
each bar is 120 microhms (including the contact resistance). In order 
to have a certain required torque and slip, the equivalent rotor resistance 
per phase of the primary circuit must be equal to 1.07 ohms. What must 
be the actual resistance of each end-ring per centimeter of its length? 
There are 100 turns per phase of the primary winding, and hi = 0.95; 
for the squirrel-cage winding k^ is always equal to unity. 

Ans. 5.8 microhms. 

46. The Circle Diagram. Let the values of primary current 
obtained from a brake test on an induction motor be plotted as 
vectors at proper phase angles with respect to the vector E\ 
of the primary voltage (Fig. 45). The locus of the ends of the 
current vectors is found to be very nearly a semicircle. This 



tti 




A/ 






X >v 






Ax. 


ls^ 


^-^^^ X \ 

^^^^^ Vy \ 


<h/^ 


fey 








flo' 









Fig. 45. The circle diagram of an induction motor. 



locus, together with some auxiliary lines, is called the circle dia- 
gram or the Heyland diagram of the induction motor. A similar 
diagram holds true for the transformer, although it is hardly ever 
used in practice. 

The importance and the convenience of the circle diagram lie 
in the fact that the complete performance of an induction motor 
can be predicted if the diameter of the semicircle and its posi- 



Chap. XIII] THE INDUCTION MOTOR 137 

tion with respect to the voltage vector Ei are known. The semi- 
circle is usually determined by the vector J of the no-load current, 
and the vector of the current I s obtained when the rotor is locked 
(the subscript s stands for starting or short circuit). At any 
other load, the extremity of the current vector 2\ lies between 
those of J and J 8 . 

The Heyland diagram is simply a graphic representation of 
the current and voltage relations in the approximately equivalent 
circuit diagram shown in Fig. 41. The exciting current and the 
no-load losses are assumed to be constant at all loads from no- 
load to standstill. The no-load current is resolved into a loss 
component J ", in phase with the voltage E h which component 
represents the iron loss, friction, and windage; and a reactive com- 
ponent Jo' which excites the main flux in the motor. At any load, 
the primary current Ji is the geometric sum of the load current 
II and the no-load current J , the vectors of these three currents 
forming a triangle. 

That the locus of the current Ji or Jl is a circle follows directly 
from eq. (212), because from it we have 

J L /sin 6 = Ei/x = const., .... (242) 

which is easily seen to be the equation of a circle in polar coordi- 
nates. The value of the constant 

E l /x = I L > (243) 

is equal to the diameter of the circle, which diameter is thus 
determined solely by the leakage reactance x of the motor. The 
smaller x is, the larger is the circle and the better the motor, 
because its power-factor is higher and its overload capacity 
larger. 

The load current I s l with the armature blocked has an energy 
component in phase with the line voltage E h because of the i 2 r 
loss in the resistances of the stator and rotor. If these resistances 
could be eliminated or put outside the motor, the load current 
on short circuit would be purely reactive and equal to Ijj = Ei/x. 
Thus, the diameter of the circle is equal in position and magni- 
tude to the load current (secondary current) of the machine with 
the armature blocked, provided that the internal resistances are 
eliminated and only leakage reactances are left. This condition 
is called an ideal short circuit. 

Leaving the loss component J " of the no-load current out of 



138 THE ELECTRIC CIRCUIT [Art. 46 

consideration, the performance of the motor is determined by 
the pure magnetizing current /</ and its ratio to the diameter II 
of the semicircle. This ratio is called the circle coefficient. Let 
the exciting reactance, or the reciprocal of b , be denoted by x . 
Then Iq = Ei/x , and, by definition, the circle coefficient 

o- = U/Il' = x/x ; '• (244) 

in other words, the circle coefficient is equal to the ratio of the 
leakage reactance to the exciting reactance. 1 

Thus, knowing the magnetizing current and the short-circuit 
current (or the magnetizing current and the circle coefficient), 
the Heyland circle can be drawn, and the relation between the 
primary current, the load current, the power-factor, and the 
angle 6 graphically determined. By drawing certain auxiliary 
lines, the input, output, slip, torque, and efficiency can also be 
read off directly from the diagram, for any assumed primary 
current. 2 In other words, the circle diagram permits one to 
determine graphically the performance characteristics, and offers 
an alternative method to the analytical procedure explained in 
Art. 43 above. The relative advantages of the analytical and 
graphical methods depend upon the problem in hand and the 
skill and temperament of the user; the student should thoroughly 
familiarize himself with both methods before deciding upon the 
use of one or the other. 

The circle coefficient is very convenient for preliminary designs 
and performance estimates. Mr. H. M. Hobart has made quite 
a study of the numerical values of this coefficient for a large num- 
ber of actually built motors, and has compiled his results in the 

1 The circle coefficient is also called the dispersion factor (Streuungs- 
koejfizient) , and is usually denoted by a. Those familiar with magnetic 
phenomena will notice that the circle coefficient is equal to the ratio of the 
permeance of the main magnetic path in the motor to that of the leakage 
paths. This is because the reactances are proportional to the corresponding 
inductances, and an inductance is equal to the permeance of the path times 
the square of the number of turns linked with it. A motor is evidently im- 
proved by reducing the permeance of its leakage paths and increasing that 
of the useful path. This means that the motor is better the lower its circle 
coefficient a. 

2 For complete and explicit instructions in regard to the construction and 
use of the circle diagram, see the author's Experimental Electrical Engineer- 
ing, Vol. 2, Chap. 29. 



Chap. XIII] THE INDUCTION MOTOR 139 

form of charts, from which the value of the coefficient may be 
taken for a motor of given or assumed dimensions. 1 

Prob. 1. Show that the circle coefficient of the motor specified in 
problem 1, Art. 43, is equal to 0.0526. 

Prob. 2. Check a few points on the curves obtained in problem 1, 
Art. 43, by constructing the circle diagram of the motor. 

47. The Analytical Determination of Performance. — Exact 
Solution. 2 The predetermination of the performance charac- 
teristics of an induction motor, explained in Arts. 43 and 46, 
is based upon the approximately equivalent diagram shown in 
Fig. 41. The exact performance characteristics are obtained by 
expressing analytically the electrical relations according to the 
correct equivalent diagram shown in Fig. 42. To be absolutely 
correct, both g and b must be varied somewhat with the load, 
because (a) the friction and windage depend upon the speed, 
(b) the iron loss is not exactly proportional to the square of the 
flux, and (c) the magnetizing current is not proportional to the 
voltage. Moreover, the friction loss ought to be separated from 
the iron loss, and subtracted from the output, instead of being 
added to the input. All these corrections make the calculations 
much more involved, and, while it is well to know about them, 
they are hardly ever justified in practice. 

In large and medium-sized motors the losses and the internal 
voltage drop are comparatively small, so that the performance 
calculated according to the exact diagram differs but little from 
that obtained with much less time and ' effort, by using the 
approximate diagram. It is only in small motors, or where ex- 
treme accuracy is required for some special reason, that the pro- 
cedure given below is justified. In very small motors, say below 
one kilowatt, the difference between the approximate and the 
correct performance is quite appreciable, because of high losses 
and a large voltage drop. 

1 H. M. Hobart, Electric Motors (1910), Chapter 21. It may be of inter- 
est to note that the correct equivalent diagram (Fig. 42) also leads to a 
circle diagram, known as the Ossanna circle. Numerous articles on this 
exact diagram will be found in the various volumes of the Elektrotechnische 
Zeitschrift and Elektrotechnik und Maschinenbau. 

2 This article may be omitted if desired, because the approximate solu- 
tion given in Arts. 43 and 46 is sufficient in a great majority of practical cases. 
However, the method used in this article is of interest to the student as 
another and somewhat different application of complex quantities. 



140 THE ELECTRIC CIRCUIT [Art. 47 

It is much more convenient to plot complete performance 
curves than to calculate the performance data for a specified out- 
put. A certain external resistance R' is assumed, such as would 
give a reasonable value of slip according to eq. (214a), and the 
performance characteristics are calculated for this value of R' '. 
Then another value of R' is assumed, and the calculations are 
repeated, and so on. For an assumed value of R' the total 
admittance between the primary terminals is calculated, using 
the general method given in Art. 28, that is, adding impedances in 
series and admittances in parallel. Knowing the total admittance, 
the primary current becomes known; then II and El are calcu- 
lated, and finally the rest of the data are obtained as in Art. 43. 
The details of the calculations are as follows: 

(1) The impedance of the load plus that of the secondary 
winding = {R' + r 2 ') + jx 2 ', 

(2) Using eqs. (121) and (122), Art. 27, find the corresponding 
admittance g 2 — jb 2 . 

(3) The total admittance between the points M and N is 
(go + g 2 ) - j(b + b 2 ). 

(4) Using eqs. (123) and (124), Art. 27, find the corresponding 
impedance tmn + J%mn- 

(5) The total impedance between the primary terminals is 
Z eg = {r M N + n) + j (xmn + Xl). 

(6) The corresponding admittance Y eq = g eq — jb eg is calcu- 
lated from eqs. (121) and (122). 

(7) The primary current is 

I 1 =E 1 Y eq . ...... (245) 

(8) The voltage across MN 

^1=^1- 7iZi = #i(l -ZiF eg ). . . . (246) 

(9) The load current is 

II — Ii — Jo = fi — EaYo, 
or, substituting the values of Ji and En from eqs. (245) and (246), 
!l = E, [Y eq (1 + FoZx) - Y ]. . . . (247) 

(10) The load voltage 

E L = I L R'. ....... (248) 



Chap. XIII] THE INDUCTION MOTOR 141 

The rest of the quantities are calculated in the same manner 
as in Arts. 43 and 44. 

In numerical work, it is convenient to take Ei along the ref- 
erence axis. Having determined the value of Y eq , computations 
are begun with the composite admittance in the brackets in 
eq. (247). Either the orthogonal expressions of the form r + jx 
or the polar expressions of the form z (cos </> + j sin 0) may be 
used, according to one's preference or familiarity with one or the 
other form. The student ought to be familiar with both forms. 
The trigonometric form is convenient for multiplication and 
division, while the Cartesian form is preferable in addition and 
subtraction. It may be advisable to use both forms in the same 
problem. 

The calculator should avoid long algebraic expressions, per- 
forming numerical operations step by step. Much time is saved 
by arranging the consecutive steps in a table, so as to repeat the 
same operations mechanically for different values of R' . An 
irregularity of the values in a column is a sure indication of a 
numerical error. 

Much time is also saved by intelligently discriminating be- 
tween the principal terms and small correction factors in an ex- 
pression. For instance, in eq. (247) Y eq is large as compared to 
Y Q and to Y eq Y Zi. It would be a waste of time to figure out 
the latter expression accurately, when, in all probability, the 
principal term will be affected only by its first significant figure. 
On the other hand, the principal term, Y eq , must be calculated 
to a degree of accuracy at least equal to that desired in the result, 
if not to a higher degree. Considerable skill, experience, and judg- 
ment are necessary to determine the proper accuracy of computa- 
tions in engineering problems. This is an art which grows by 
intelligent exercise, and it is never too early to begin practicing 
it. The rewards are time and mental energy saved for better 
things, while obtaining an accuracy which is commensurate with 
the desired result. 1 

1 For a complete set of final formulae for induction motor characteristics, 
see Arnold's Wechselstromtechnik, Vol. 5, part 1 (1909), pp. 65-78. A very 
slight inaccuracy is introduced there in the beginning, by neglecting the 
imaginary part in a complex quantity. See also Dr. Steinmetz's Alternating- 
current Phenomena, under "Induction Motor." 



142 THE ELECTRIC CIRCUIT [Art. 47 

Prob.-l. Make out a table showing in detail the order of computa- 
tions for a complete set of performance characteristics of an induction 
motor, according to the method developed above. 

Prob. 2. Mark on the curve sheet obtained in problem 1, Art. 43, a 
few points determined according to the exact equivalent diagram, in 
order to see the inaccuracy resulting from the use of the approximate 
method. 



CHAPTER XIV 

THE DIELECTRIC CIRCUIT 

48. The Electrostatic Field. 1 In the following discussion, it 
is assumed that the student knows the fundamental phenomena 
of electrostatics from his study of physics. The purpose of the 
treatment given here is to deduce the principal numerical relations 
which are of importance in electrical engineering. The electro- 
static field is considered in this book from Faraday's point of 
view, viz., as consisting of displacements of electricity, and stresses 



Battery 




Ball, f J 
Gv.(/ 



Battery 



Condenser 





K^ * 


♦ 


— *->. 


'-N s \ 1 


""? 


":>' 


% 


^B 


:~ 


^Q 


M. 





Fig. 46. A plate condenser completing a direct- current circuit. 

in the dielectric. This is different from the older theory of the 
action of electric charges at a distance. 

Let a source E of continuous electromotive force (Fig. 46) be 
connected to two parallel metallic plates A and B, the combina- 
tion of which is commonly known as a condenser. Let the plates 

1 See the footnote at the beginning of Chapter 3. 
143 



144 THE ELECTRIC CIRCUIT [Art. 48 

be separated from each other by air, or by some other non-conduct- 
ing material. When the key K is pressed upwards, a certain 
quantity of electricity, Q, flows from the battery to the plate A, 
and the same quantity flows from the plate B back to the battery. 
This quantity can be measured by the ballistic galvanometer 
shown in the circuit. Within a very short time the difference of 
potential between the plates becomes equal and opposite to that 
of the battery and the flow of current stops. 

Since electricity behaves like an incompressible fluid, the same 
quantity, Q, is displaced through the whole circuit, including the 
layer of insulation or dielectric between the condenser plates. 
This displacement is accompanied by a stress in the dielectric, 
similar in some respects to a mechanical stress in an elastic body. 
The directions of the electric stress and of the lines of displacement 
of electricity through the air are shown in the figure by dotted 
lines. These stresses produce a counter-electromotive force, 
which finally balances that of the battery. When the key is 
opened, the condenser remains charged, since the stress and the 
displacement can be relieved only in a closed circuit. To dis- 
charge the condenser, its plates must be connected by a conductor; 
this is done by pressing the key down. The deflection of the 
ballistic galvanometer during the discharge is equal and opposite 
to that during the charge, and the electric energy stored in the 
condenser is dissipated by the current in the form of heat. 

The difference between a dielectric and a conductor is that the 
resistance of the former to the passage of electricity is of an elastic 
nature; that is, the stress can be relieved and the stored energy 
returned to the circuit. On the contrary, the resistance to the 
flow of electricity in a conductor is of the nature of friction. 
The energy is converted into Joulean heat and cannot be restored. 

The modern electronic theory of electricity is not sufficiently 
advanced at this writing to give a clear account of the true nature 
of these displacements and stresses in a dielectric. It is therefore 
preferable for our purposes not to specify the mechanism by which 
these stresses and displacements are produced. We shall simply 
assume, as a matter of fact, the structure of dielectrics to be such 
that an e.m.f. across a layer of such material produces a displace- 
ment of a certain quantity of electricity, which is proportional to 
the e.m.f. When the e.m.f. is removed and a closed circuit is 
provided, the stresses within the dielectric are relieved, and the 



Chap. XIV] THE DIELECTRIC CIRCUIT 145 

displacement disappears. The analogy to an elastic body sub- 
jected to external mechanical forces naturally suggests itself. 

Experiment shows that, with given metallic plates (Fig. 46) 
and the same applied e.m.f., the value of the electric displacement 
depends upon the nature of the dielectric. With solid and liquid 
insulating materials, such as glass, oil, mica, etc., the same e.m.f. 
produces larger displacements of electricity than with air as the 
dielectric. These materials are therefore said to possess higher 
'permittivity than the air (some writers use the word inductivity). 

When an alternating voltage is applied at the terminals of a 
condenser, the displacement of electricity in the dielectric varies 
continually in its magnitude and periodically reverses its direc- 
tion; consequently, it gives rise to an alternating current in the 
conducting part of the circuit. This is called the charging or 
capacity current. This current leads the alternating voltage in 
phase by 90 degrees, as may be seen from the following consider- 
ations : When the voltage has reached its instantaneous maximum 
the charging current is zero, because at the crest of the wave the 
voltage and the displacement remain practically constant for a 
short period of time. As soon as the voltage begins to decrease, 
the current begins to flow in the direction opposite to that of the 
applied voltage, because the elastic reaction of the dielectric is 
now larger than the applied electromotive force. At any instant, 
the current, or the rate of flow of electricity, is proportional to 
the rate of change of the applied voltage. But if the applied 
voltage varies according to the sine law, the rate of variation is 
also represented by a sine function differing in phase by 90 de- 
grees from the original function, because d (sin.x)/dx = cos x = 
sin (90°+#); see also Art. 66 below. That there must be a dis- 
placement of 90 degrees between the voltage and the current 
follows also directly from the assumed elastic structure of the 
dielectric. The energy is supposed to be periodically stored in 
the dielectric and given up again without any loss; hence, the 
average power must be zero, and the current must be reactive. 

49. A Hydraulic Analogue to the Dielectric Circuit. The 
hydraulic analogue shown in Fig. 47 may assist the student in the 
understanding of the electrostatic circuit. A is a pump which 
corresponds to the source of electromotive force in Fig. 46. The 
pipes B and C represent the leads to the condenser, or the metallic 
parts of the circuit. The cylinder D corresponds to the condenser, 



146 



THE ELECTRIC CIRCUIT 



[Art. 49 



and the elastic partition K is analogous to the dielectric. Let 
the pipes and the cylinders be filled with water, and let the piston 
in A be in its middle position, the partition K not being stressed. 
Let the stopcock M be open, and the stopcock N closed. When a 



I Piston Rod 



Water 
Meter 




Fig. 47. A hydraulic analogue of a dielectric circuit. 



pull to the right is exerted upon the piston rod and it is forced 
to move, the water in the system is displaced, and the elastic 
partition K is strained, as shown in the figure. With a given 
pull, or a given electromotive force, the movement stops when 
the pull is balanced by the elastic reaction of the partition. 
The charge, or the total displacement, is represented by the 
amount of water shifted; it can be measured by the water-meter 
W, which thus takes the place of the ballistic galvanometer. 

If the pipes are frictionless, and the inertia of the piston and 
water is assumed negligible, the analogy can be followed still 
further; namely, the phase difference in time between the pull 
and the velocity of the water is equal to 90 degrees, the velocity 
leading the pull. Assuming the motion of the piston to be har- 
monic, the velocity of the flow of water is at its maximum when 
the piston is at the center of its stroke. The required pull is equal 
to zero at this moment, because the elastic partition is in its 
middle, or unstrained position. At the end of the stroke the 
velocity is zero, but the pull is at its maximum, because the 
partition is strained to its extreme position, and exerts its maxi- 
mum elastic reaction. Thus the pull lags behind the velocity. 

Substituting another partition, made of a more yielding 



Chap. XIV] THE DIELECTRIC CIRCUIT 147 

material (material possessing higher permittivity), a larger dis- 
placement is produced with the same pull; this corresponds to 
the case in which some solid or liquid dielectric is substituted for 
the air. 

Closing the stopcock M corresponds to breaking the electric 
circuit of the condenser. It will be seen from analogy that the 
condenser remains charged. To discharge the condenser, the 
stopcock N must be opened; this equalizes the pressure on both 
sides of the elastic partition. Since, in reality, water possesses 
some inertia, the partition does not stop in its middle position 
during the discharge, but the momentum of the water carries it 
beyond the center. The electromagnetic inertia of the electric 
current produces a similar effect, and we thus have a simple 
explanation of the oscillatory character of the electric discharge. 
During this discharge, the energy is alternately transformed into 
the potential energy of dielectric stress, and into kinetic energy of 
the magnetic field. The oscillations of the partition are gradually 
damped out by the frictional resistance of the pipes. In the 
electric circuit, oscillations are damped by the ohmic resistance 
of the conducting parts of the circuit. 

The student can follow this analogy still further, and explain 
free electrical vibrations, current and voltage resonance, also the 
effect of a resistance in series and in parallel with a condenser, etc. 

50. The Permittance and Elastance of Dielectric Paths. 
Let Q (Fig. 46) be the total displacement of electricity in the 
dielectric, measured in ampere-seconds or coulombs, and let E 
be the voltage impressed across the condenser or " permittor." 
Experiment shows that up to a certain limit Q is proportional to 
E; this is similar to the behavior of an elastic body, in which the 
strains are proportional to the applied forces until the limit of 
elasticity has been reached. Thus, we may write 

Q = CE, . (249) 

where the coefficient of proportionality, C, is called the permittance 
of the condenser. The older name for C is electrostatic capacity. 
When E is in volts and Q in coulombs, permittance is measured 
in units called farads. A condenser has a permittance of one 
farad when a displacement of one coulomb is produced for each 
volt applied at its terminals. The farad being too large a unit 
for practical use, permittances are usually measured in micro- 



148 THE ELECTRIC CIRCUIT [Art. 50 

farads, one microfarad being equal to one millionth part of a 
farad. 

The larger the permittance of a condenser, the larger is the 
displacement of electricity with the same voltage; hence C is a 
measure of the ease with which an electric displacement can be 
produced in a given condenser. In this respect the concept of 
permittance is analogous to those of electric conductance and 
magnetic permeance. 

In some cases it is convenient to speak, not of the degree of 
ease, but of the difficulty with which an electric displacement 
can be produced in a given condenser. For this purpose, a 
coefficient of proportionality, the reciprocal of C, has to be used; 
and eq. (249) becomes 



where 



E = SQ, (250) 

8 = C- 1 (251) 



is called the elastance of the condenser. Elastance is thus analo- 
gous to electric resistance and to magnetic reluctance. When 
permittance is measured in farads, the unit of elastance is the 
reciprocal of the farad, and may therefore be properly called the 
daraf. This is a name derived by spelling the word farad back- 
wards, that is, in the same way in which mho is derived from ohm. 1 
A condenser has an elastance of one daraf when one volt of pres- 
sure is required for each coulomb of displacement within it. The 
farad being too large a unit for practical use, the daraf is con- 
sequently too small a unit. Therefore, in practice, elastances 
should be measured in megadarafs, one megadaraf (= 10 6 daraf s) 
being the reciprocal of one microfarad. 2 

When two or more permittances are connected electrically 
in parallel, the resultant permittance is larger than that of any 
of the component condensers, because a larger path is offered to 

1 It may be of interest to mention in this connection a similar derivation 
of the name for a unit of magnetic reluctance. The henry being the natural 
unit of magnetic permeance (or inductance) in the ampere-ohm system, the 
author has proposed calling the corresponding unit of reluctance the yrneh, a 
word derived by spelling the word henry backwards. See his Magnetic Circuit, 
Art. 5. 

2 For a complete rational nomenclature of electric and magnetic quanti- 
ties, see the table on page xii at the beginning of the book, and also the one in 
the Appendix. 



Chap. XIV] THE DIELECTRIC CIRCUIT 149 

the displacement. The relation is similar to that of conductances 
or permeances in parallel. Let C h C 2 , etc., represent permit- 
tances connected in parallel across a source of constant voltage 
E, and let Qi, Q2, etc., be the corresponding electric displacements 
through these condensers (or permittors). Then, according to 
the definition of permittance, we have 

Qi = dE 

Q2 = C 2 E (252 



The equivalent permittance, C eq , must be such as to allow of a 
displacement equal to the sum of the partial displacements, with 
the same voltage; hence, 

SQ = C eq E (253) 

Adding eqs. (252) together gives 

Q1 + Q2 + etc. = E (Ci + C 2 + etc.), 
or, by comparison with eq. (253), 

C eq = 2C (254) 

In other words, when permittances are connected in parallel, 
the equivalent permittance is equal to their sum. 

When condensers (or elastors) are connected in series, it is 
more convenient to use their elastances. Since electricity behaves 
like an incompressible fluid, the displacement through several 
elastances in series is the same in all of them. Let this displace- 
ment be denoted by Q, and let the voltages across the terminals 
of the individual elastors be E h E 2 , etc. Then, 

E 1 = SiQ 

E 2 = S 2 Q 



where Si, $2, etc., are the elastances of the separate condensers. 
The equivalent elastance must allow of the same displacement Q 
with the same total voltage, or 

2E = S eq Q (256) 

Adding eqs. (255) together gives 

Ei + E 2 + etc. = Q (Si + S 2 + etc.), 



150 THE ELECTRIC CIRCUIT [Art. 51 

or, by comparison with eq. (256), 

S eq = 2S (257) 

In other words, when elastances are connected in series, the 
equivalent elastance is equal to their sum. The analogy to the 
addition of conductances in parallel and resistances in series is 
self-evident (see Art. 3). 

Prob. 1. A condenser, which has a permittance of 10 microfarads, is 
connected to a direct-current magneto, the speed of which is increased at 
a uniform rate, so that the voltage rises at a rate of 1.7 volts per second. 
Calculate the charging current. 

Ans. 17 microamperes. Note: This is the principle of an appara- 
tus used for measuring the acceleration of railway trains. 

Prob. 2. An elastance of 10 kilodarafs is connected across a 220- 
volt, 50-cycle line. Show that the effective value of the charging current 
is 6.91 amp. Solution: The maximum displacement in the dielectric is 
220 V2/(10 X 10 3 ) = 22 V2 X 10~ 3 coulombs. This displacement is 
reduced to zero within zlo of a second; hence, the average charging 
current is 4.4 V2 amp. The effective value, assuming a sine- wave of 
current, is 4.4 V2 X (| ir/V2) = 6.91 amp. 

Prob. 3. Show that with two condensers in parallel the ratio of the 
displacements equals that of the permittances or is inversely as the ratio 
of the elastances. What is the analogous relation for conductances and 
resistances? 

Prob. 4. When two condensers are in series, show that the ratio of 
the voltage drops across them equals that of the elastances, or is inversely 
as the ratio of the permittances. What is the analogous relation for 
resistances and conductances? 

Prob. 5. A sectionalized condenser, such as is used for calibration 
and exact measurements, is built up of the following permittances: 
0.5, 0.2, 0.2, 0.05, and 0.05 microfarads. What is the extreme range 
of permittances and elastances possible by combining these sections in 
series and in parallel? 

Ans. From 1 to 0.0192 mf., or from 1 to 52 mgd. 
Prob. 6. Referring to the preceding problem, the sections of the con- 
denser are connected as follows: 0.2, 0.05, and 0.05 mf. are in series, and 
the combination is shunted by 0.2 mf. Then the whole is put in series 
with 0.5 mf. Show that the resultant permittance is equal to 0.154 
microfarads. 

51. Permittivity and Elastivity of Dielectrics. Experiment 
shows that the permittance of a sample of any dielectric varies 
with its dimensions in the same way that the conductance of a 
metal or the permeance of a magnetic path in a non-ferrous medium 
does; namely, the permittance is proportional to the cross-sec- 
tion of the layer and inversely proportional to its length in the 



Chap. XIV] THE DIELECTRIC CIRCUIT 151 

direction of the lines of force. By increasing the cross-section of 
the path perpendicular to the lines of force (Fig. 46) , the displace- 
ment is increased in the same proportion. On the other hand, 
the displacement is found to be inversely proportional to the 
thickness of the dielectric, since the distance through which the 
voltage must act is greater if the thickness is increased. These 
relations follow directly from the laws deduced in the preceding 
article for the addition of permittances in parallel and elastances 
in series. Thus, by analogy with eq. (21), Art. 5, we put 

C = kA/1, (258) 

where k is called the permittivity of the dielectric. It is analogous 
to the conductivity of a conducting material, or to the perme- 
ability of a magnetic medium. Permittivity may be defined as 
the permittance of a cubic unit of dielectric, when the lines of 
displacement are straight lines perpendicular to one of its faces. 
For air the permittivity is 

Ka = 0.08842 X 10~ 6 microfarads per cm. cube. . (259) 
For other dielectrics, liquid and solid, the permittivity is higher than 
that of air; that is to say, they are more yielding to an electro- 
motive force. It is convenient to express their permittivities in 
terms of that of the air; for instance, we may say that the per- 
mittivity of a certain transformer oil is 2.1 times that of the air. 
The relative permittivities of some important insulating ma- 
terials are tabulated in Art. 56 below, merely to indicate their 
order of magnitude. For accurate values, the reader is referred 
to various published physical tables and engineering handbooks. 
The older name for relative permittivity is specific inductive 
capacity (or dielectric constant). It is more convenient in prac- 
tice to use relative than absolute permittivities, because the 
necessity of tabulating small quantities like K a in eq. (259) is 
avoided. Besides, the data are more readily comparable with 
one another, and with the permittivity of air, which is a standard 
dielectric. This procedure is analogous to tabulating the con- 
ductivities of various metals in terms of that of pure copper, 
taken as 100 per cent. The absolute permittivity of a material 
is obtained by multiplying the absolute permittivity of air by the 
relative permittivity of the dielectric in question. Equation (258) 
thus becomes 

C = KK a A/l, ....... (260) 

where K stands for the relative permittivity. 



152 THE ELECTRIC CIRCUIT [Art. 51 

The elastance of a prismatic piece of dielectric, with the lines 
of displacement parallel to one set of its edges, is expressed by 
analogy with eq. (20), Art. 5, as 

S = all A, (261) 

where 

<t = k- 1 (262) 

is called the elastivity of the dielectric. Elastivity is analogous to 
the resistivity of a conducting material or to the reluctivity of a 
magnetic medium, and may be expressed for practical purposes in 
megadarafs per centimeter cube. For air, the absolute elastivity 
is, according to eq. (259), 

(T a = K a ~ l = 11.3 X 10 6 megadarafs per cm. cube. . (263) 
The concept of relative elastivity could be introduced if necessary, 
in which case its values would be equal to the reciprocals of the 
relative permittivities tabulated in Art. 56. However, it is suf- 
ficient to use the relative permittivity, even when dealing with 
elastances, so that eq. (261) becomes 

S= (<r a /K)l/A. (264) 

The nomenclature used above is due to Mr. Heaviside; 1 it is 
consistent and uniform with the nomenclature used in the electro- 
conducting and magnetic circuits, and is suggestive as to the nature 
of the phenomena. The electrostatic nomenclature now in general 
use comprises but three terms; namely, condenser, capacity, and 
specific inductive capacity. It is hoped that the more rational and 
complete nomenclature used here will help to a clearer understand- 
ing of the dielectric circuit, and will simplify engineering calcula- 
tions relating thereto. 2 

Note: The author considers the above-given value of n a , eq. (259), 
to be an experimental coefficient, in the same sense in which other prop- 
erties of materials are characterized by experimental coefficients. For an 
engineer, the volt and the ampere are arbitrary units established by an 
international agreement, no matter what their relation to the so-called 
absolute units. The value of n a can be calculated theoretically, assuming 
the ratio between the electrostatic and the electromagnetic units to be 
known. In the absolute electrostatic system of units, with air as the 
dielectric, a plate condenser having an area of A sq. cm. and a distance 
between the plates equal to I cm., has a capacity equal to A/(M). The 

1 O. Heaviside, Electromagnetic Theory (1894), Vol. 1, p. 28. 

2 See the author's paper "Sur Quelques Calculs Pratiques des Champs 
Electrostatiques," in the Transactions of the Congresso Internazionale delle 
Applicazioni Elettriche, Turin, 1911. 



Chap. XIV] THE DIELECTRIC CIRCUIT 153 

factor 4 tt enters on account of an unfortunate selection of the expression 
for Coulomb's law, which should have been <?ig 2 /4 -n-r 2 , instead of qiq 2 /r 2 . 
In the absolute electromagnetic units the same capacity is equal to 
(A/4 wl) (3 X 10 10 ) -2 , where 3 X 10 10 is the velocity of light in centimeters 
per second. To obtain the result in microfarads, the foregoing expression 
must be multiplied by 10 15 . On the other hand, the same capacity 
expressed in the rational units defined above is k g A /I. Equating the two 
expressions gives K a = 10 -5 /(9 X 4 tt) = 0.08842 X 10 -6 microfarads per 
centimeter cube. 

The fact that k (1 can be expressed through the velocity of light does 
not make n a the less an empirical coefficient, because the velocity of light 
itself is determined experimentally. As a matter of fact, one of the ways 
in which the velocity of light is determined consists in calculating it 
indirectly from the value of n a obtained from measurements. 

Prob. 1. Show that in the English system K a = 0.2244 X 10 -6 micro- 
farads per inch cube. 

Prob. 2. A condenser (Fig. 46) consists of two metal plates, 50 by 
70 cm. each, in contact with a glass plate 3 mm. thick between them. 
When a continuous voltage of 2400 is applied to the condenser, the ballis- 
tic galvanometer shows a charge of 17.1 microcoulombs. What is the 
relative permittivity of the glass? Ans. 6.9 

Prob. 3. A 0.5-mf. mica condenser is to be made out of sheets of mica 
12 by 25 cm., 0.3 mm. thick, and coated on one side with a very thin 
film of silver. How many sheets are required? The relative permittivity 
of the mica is about 6. 

Ans. About 96 sheets, 48 sheets in parallel per terminal. 

Prob. 4. Let the dielectric in problem 2 consist, instead of glass, of 
three layers of different materials. Let the thicknesses of these layers 
be 1.2, 0.7, and 1.1 mm., and let the corresponding values of relative per- 
mittivities be 2, 3, and 5. What is the capacity of the condenser? Hint : 
Calculate the equivalent elastance as the sum of three elastances in series. 

Ans. 2.94 X 10~ 3 mf. 

52. Dielectric Flux Density and Electrostatic Stress (Voltage 
Gradient). Referring again to the uniform electrostatic field 
(Fig. 46), consider a cube of the dielectric, one square centimeter 
in cross-section, and one centimeter long in the direction of the 
lines of force. Let a quantity of electricity Q be supplied by the 
battery, as shown by the ballistic galvanometer; then the same 
quantity of electricity must be displaced in the dielectric. Neg- 
lecting a small displacement at the edges and at the outside sur- 
faces of the plates, the whole quantity Q is uniformly displaced 
between the plates. Therefore, if the area of each plate is equal 
to A square centimeters, the displacement through the cube under 
consideration is equal to 

D = Q/A (265) 



154 THE ELECTRIC CIRCUIT [Art. 52 

Since Q is the total electrostatic flux, D is naturally called the 
dielectric flux density. If Q is measured in coulombs, D is ex- 
pressed in coulombs per square centimeter. In practice, Q is 
measured in microcoulombs, and D is expressed in micro coulombs 
per square centimeter. The dielectric flux density is analogous 
to current density U (Art. 6) and to magnetic flux density B. 

When an electrostatic field is non-uniform (Fig. 48), it is con- 
veniently subdivided by lines of force and equipotential surfaces 
perpendicular to the same. The procedure is similar to that used 
in Art. 8. In this case, the total flux or displacement divided by 
the area of an equipotential surface gives only the average flux 
density through the surface. The actual density varies from 
point to point, and it is therefore proper to speak of the dielectric 
flux density at a point. Take a tube of infinitesimal cross-section 
formed by lines of force, and let dQ be the displacement of electric- 
ity through this tube. The displacement is the same through 
any normal cross-section of the tube, because electricity behaves 
like an incompressible fluid. Let dA be a particular cross-section 
of the tube; then the flux density at this cross-section is 

D = dQ/dA, (266) 

D being usually expressed in coulombs (or microcoulombs) per 
square centimeter. Since the cross-section of the tube is infinites- 
imal, D is the density at the point corresponding to the position 
of dA. 

If the flux density in a uniform field is given, the total displace- 
ment is 

Q = DA. ...... . (267) 

In a non-uniform field, the flux density must be given as a function 
of the coordinates of the field; so that 

Q = C A D dA, (268) 

the integration being extended over the whole area of an equi- 
potential surface, or over the part of this area through which the 
flux is. to be calculated. 

The electromotive force impressed at the terminals of a 
condenser is balanced in the whole thickness of the dielectric; 
that is, each small length of path in the dielectric produces its 
own counter-electromotive force. Therefore, it is possible to 
speak of the voltage drop per unit length of the path in the 



Chap. XIV] 



THE DIELECTRIC CIRCUIT 



155 



dielectric, the same as in Art. 6. This voltage gradient, or electric 
intensity, in a uniform field is expressed by 

G = E/l, (269) 

and is measured, as in the conducting circuit, in volts per centi- 
meter, kilovolts per millimeter, or in other suitable units. 

In a non-uniform field, the electric intensity, or voltage gradi- 
ent, varies from point to point. Let the voltage between two 
infinitely close equipotential surfaces MN and M'N' (Fig. 48) 




Fig. 48. A non-uniform electrostatic field, represented by lines of 
displacement and equipotential surfaces. 

be dE, and let the distance mn between the surfaces, along a 
certain line of force HH f , be dl. Then the voltage gradient along 
mn is 

G = dE/dl . (270) 

The length of the line mn being infinitesimal, G is the intensity at 
any point between m and n. 

When the voltage gradient is uniform, we have for the total 
voltage across the field 

E = Gl . . (271) 

In a non-uniform field, G has to be given as a function of I, so that 

E = f l Gdl, (272) 



156 THE ELECTRIC CIRCUIT [Art. 52 

the integration being performed between any two points on the 
equipotential surfaces between which the voltage is to be deter- 
mined. 

Imagine a uniform field existing in a dielectric, and consider a 
unit cube of the material. The total displacement through such 
a cube is equal to the flux density D, and the voltage across it is 
equal to the voltage gradient G. The permittance and the elas- 
tance of the cube are respectively equal to the permittivity and 
the elastivity of the material. Thus, applying to the cube 
eqs. (249) and (250), we have 

D = kG, (273) 

and 

G = aD (274) 

These equations are analogous to eq. (25) in Art. 6 of this book 
and to eqs. (15) and (16), Art. 8, of the Magnetic Circuit. These 
relations may be considered as fundamental in the theory of the 
dielectric circuit, G being the cause, D the effect, and k (or a) 
the coefficient of proportionality which characterizes the material. 
Similar linear relations between cause and effect hold in the con- 
duction of heat, and in the theory of elasticity. The voltage 
gradient is sometimes called the stress in the dielectric, eq. (274) being 
analogous to Hooke's law for elastic bodies. The elastivity a takes 
the place of the modulus of elasticity. 

If the field or the dielectric is non-uniform, eqs. (273) and (274) 
still hold true for every point, D being the dielectric flux density 
and G the voltage gradient at the point considered. This can be 
proved by applying eqs. (249) and (250) to an infinitesimal par- 
allelopiped instead of a unit cube. 

Equations (268) and (272) are expressed in words by saying 
that the total displacement Q is the surface integral of the dielectric 
flux density D, and the voltage E is a line integral of the gradient 
or stress G. These statements are almost self-evident from the 
definition of the quantities and the structure of dielectrics. 

Prob. 1. What are the dielectric flux density and voltage gradient in 
problem 2, Art. 51? Ans. 4.885 X 10~ 3 mc./cm. 2 ; 8 kv./cm. 

Prob. 2. The condenser specified in problem 4, Art. 51, is subjected 
to a difference of potential of 10 kv. What are the voltage gradients 
(stresses) in the three layers of dielectric? 

Ans. 4.75; 3.16; 1.9 kv./mm. 



CHAPTER XV 
THE DIELECTRIC CIRCUIT — (Continued) 

53. Energy in the Electrostatic Field. When a dielectric 
is being charged, a current flows into it from the source of elec- 
tromotive force. This involves the expenditure of a certain 
amount of energy, because the counter-e.m.f. due to the dielectric 
stresses has to be overcome. This energy is not converted into 
heat, and lost, as in the case of metallic conduction: it is stored 
in the dielectric in potential form, and can be returned to the cir- 
cuit by reducing the voltage at the condenser terminals. With 
reference to the analogy shown in Fig. 47, the mechanical energy 
expended by the pump in straining the elastic partition is stored 
in the partition, in the form of potential energy. This energy 
can be returned to the piston rod by allowing it to be moved by 
the elastic forces of the partition. 

In some cases it is necessary to calculate the energy stored in 
an electrostatic field; or to express the energy stored per cubic 
centimeter of dielectric, as a function of the stress G and flux 
density D, at the point under consideration. 

Consider first the simple case of a uniform field (Fig. 46) , 
and neglect the small amount of displacement occurring outside 
the space between the plates. Let the dielectric be charged by 
gradually raising the voltage between its limiting surfaces from 
zero to a final value E; and let e and i be the instantaneous values 
of the voltage and charging current at a moment t during the 
process of charging. 1 The total electrical energy delivered to the 
dielectric in charging it is 

W= f eidt = f e-dq, .... (275) 
Jo Jo 

where T is the total time of charging, and dq = i dt is the infini- 
tesimal charge or displacement added to the condenser during 
the interval of time dt. The quantities dq and e can be expressed 

1 The voltage and the charging current rise gradually, even though the 
key K be closed suddenly. This is on account of an ever-present magnetic 
inductance which acts as a kind of electromagnetic inertia. 

157 



158 THE ELECTRIC CIRCUIT [Art. 53 

through the instantaneous flux density D t and the stress G t ; 
namely, from eq. (267), dq = A dD t , and from eq. (271) e = G t l. 
Performing the substitution, and taking the constant quantities A 
and I outside of the sign of integration, we get 

W = Al C T G t dD t (276) 

In order to integrate this expression, D t must be expressed through 
G t , or vice versa. The relation between the two is given by eq. 
(273). Eliminating D t , we obtain 

W = kAI f T G t dG t = i kVG 2 , . . . (277) 

where V = Al is the volume of the dielectric, and G is the final 
value of the stress, at the time T. Hence, the energy stored per 
unit volume of the dielectric, or the density of energy, is 

W' = W/V = § kG 2 = i G 2 /a "(278) 

Using relations (273) and (274), the preceding formula can also 
be written in the following forms : 

W ' = \ GD = \ D 2 /k = J aD 2 (279) 

The analogy to the corresponding formulae in Art. 69 of the Mag- 
netic Circuit is apparent at once. 

The total stored energy can be expressed through the per- 
mittance or elastance of the dielectric. We have from eq, (249) 
dq = C *de; substituting in eq. (275) and integrating, we get 

W = | CE* = J EyS. ..... (280) 

Since the final charge, or total displacement Q equals CE or 
E/S, the energy can be represented also in the following forms: 
W=iQE = iQ*/C = iQ*S (281) 

These formulae are analogous to the corresponding expressions in 
Art. 57 of the Magnetic Circuit. 

Let now the dielectric and the field be of an irregular form 
as shown in Fig. 48. The stress G and the displacement D are 
different at different points, so that it is necessary to consider 
infinitesimal layers of the dielectric between consecutive equi- 
potential surfaces, and infinitesimal threads of displacement be- 
tween .the electrodes. Consider an infinitesimal volume mnqp of 
the dielectric, comprising the part of a tube of displacement HH' 
between two equipotential surfaces MN and M'N'. The sides 



Chap. XV] THE DIELECTRIC CIRCUIT 159 

mp and nq can be provided with infinitely thin metal films, because 
these sides lie in the equipotential surfaces, and therefore no 
current would flow along these metal coatings. Then the element 
of volume under consideration is converted into a small plate 
condenser; the flux density and the stress within this element can 
be considered as uniform, so that formula (277) holds true, and 
we have 

dW = $KG*-dV (282) 

Differentials are used because both the volume and the stored 
energy are infinitesimal. The density of energy 

W' = dW/dV = \ kG 2 , (283) 

and has the same expression as in the case of a uniform field; but 
its numerical value is different from point to point, because G is 
variable. The other expressions for the density of energy, eqs. 
(278) and (279), also hold true for the points of a non-uniformly 
stressed dielectric, provided that proper values of D and G are 
used for each point. 

The total energy stored in a non-uniform electrostatic field is 

W = | f V KG 2 dV = | f V GDdV = | f D 2 dV/K; (284) 
Jo Jo Jo 

two more expressions may be written in which l/a is used in place 
of k. In order to perform the integration G and D must be given 
as functions of coordinates, and the integration extended over 
the whole space occupied by the field. Equations (280) and (281) 
are true for condensers of any shape, because in the deduction of 
these formulae no assumption is made as to the particular form of 
the dielectric or the electrodes. 

The expressions for the electrostatic energy of the field, 
derived above, are aualogous to the corresponding ones for the 
potential energy of stressed elastic bodies; and this is consistent 
with the assumed behavior of dielectrics. Consider the work 
necessary per cubic centimeter to strain mechanically the elastic 
fibers of a given material. The external mechanical force being 
applied gradually (so as to avoid oscillations), the stress varies 
from zero to its final value G. Let G t be some intermediate value 
of the stress, and let D t be the corresponding strain. The same 
symbols G and D are used here to denote the mechanical quantities 
analogous to electric stress and displacement. While the strain 



160 THE ELECTRIC CIRCUIT [Art. 54 

increases from D t to (D t + dD t ), the stress G t may be considered 
constant; the infinitesimal work done is therefore equal to G t dD t . 
The total work of deformation is 



-j> 



W = G t dD t . 

But, according to Hooke's law of elasticity, strains are propor- 
tional to stresses, so that a linear relation exists between D t and 
G t , similar to eq. (274). We thus arrive again at the result that 
the work necessary to strain one cubic unit of an elastic material 
is equal to J aD 2 . 

Prob. Calculate the total stored energy, and the density of energy, in 
the condenser given in problem 2, Art. 51. 

Ans. 20.52 milliwatt-seconds (millijoules); 19.53 micro joules per 
cubic centimeter. 

54. The Permittance and Elastance of Irregular Paths. 1 In 

most practical cases where it is required to determine the per- 
mittance or elastance of a dielectric, for instance in high-tension 
apparatus, the geometric shapes of the metal parts and of the 
insulation are either irregular or too complicated to be expressed 
analytically. It is therefore necessary in such cases to determine 
the shape of the field by trials and approximations, or by experi- 
ment.. The general law, substantiated by all known experiments, 
is as follows : The distribution of the lines of force and equipotential 
surfaces in a dielectric is such as to make the total permittance a maxi- 
mum, or the elastance a minimum. 

This is a particular case of the general law of nature known as 
the law of minimum resistance. Let a condenser of irregular 
shape (Fig. 48) be connected to a source of unlimited energy, 
having a constant voltage E. The law of minimum resistance 
requires that the dielectric take in as much energy as is compati- 
ble with its properties. This means that expression (280) must 
be a maximum; that is, with constant E, the permittance C must 
be a maximum, or the elastance S a minimum. 

Now let. it be required to establish a given flux in a certain 
dielectric; in other words, let Q be a constant. The law of mini- 
mum resistance requires in this case that the result be accom- 

1 The treatment is similar to that of conductors of irregular shape, given 
in Art. 10 of this book, and of irregular magnetic paths in Art, 41 of the 
Magnetic Circuit. 



Chap. XV] THE DIELECTRIC CIRCUIT 161 

plished with the least possible expenditure of energy. According 
to eq. (281), we have again the same condition of maximum C or 
minimum S. 

Therefore, in order to calculate the permittance (or the elas- 
tance) of a given dielectric, or to find the flux densities and stresses 
in different parts of it, proceed as follows: The field is mapped 
out into small cells by lines of force and equipotential surfaces, 
drawing them to the best of one's judgment; the total permittance 
is calculated by properly combining the permittances of the cells 
in series and in parallel. Then the assumed directions are some- 
what modified, the permittance is calculated again, and so on; 
until by successive trials the positions of the lines of force are 
found with which the permittance becomes a maximum. 

The work of trials is made more systematic by following a 
procedure suggested by Lord Rayleigh. Imagine infinitely thin 
sheets of metal (material of infinite permittivity) to be interposed 
at intervals into the field under consideration, in positions approxi- 
mately coinciding with the equipotential surfaces. If these sheets 
exactly coincided with the actual equipotential surfaces, the total 
permittance of the field would not be changed, there being no 
tendency for the flux to pass along the equipotential surfaces. In 
any other position of the conducting sheets, the total permittance 
of the field is evidently increased. Moreover, these sheets become 
new equipotential surfaces of the system, because no difference of 
potential can be maintained along a path of infinite permittance. 
Thus, by drawing in the given field a system of surfaces approxi- 
mately in the directions of the true equipotential surfaces, and 
assuming these arbitrary surfaces to be the true ones, the true 
elastance of the path is reduced. In other words, by calculating 
the elastances of the laminae between the " incorrect " equipotential 
surfaces and adding these elastances in series, one obtains an 
elastance which is lower than the true elastance of the field. This 
gives a lower limit for the required elastance (or an upper limit 
for the permittance) of the field. 

Imagine now the various tubes of force of the original field 
wrapped in infinitely thin sheets of a material of zero permittivity 
or infinite elastivity (absolute insulator). This does not change 
the elastance of the paths, because no flux passes between the 
tubes. But if these wrappings are not exactly in the direction 
of the lines of force, the elastance of the field is increased, because 



162 THE ELECTRIC CIRCUIT [Art. 54 

the insulating wrappings displace the lines of force from their 
natural positions. Thus, by drawing in a given field a system of 
surfaces approximately in the directions of the lines of force, 
calculating the permittances of the individual tubes, and adding 
them in parallel, an elastance is obtained which is higher than the 
true elastance of the field. This gives an upper limit for the 
elastance (or a lower limit for the permittance) of the path under 
consideration. 

Therefore, the practical procedure is as follows: Divide the 
field to the best of your judgment into cells, by equipotential 
surfaces and tubes of force, and calculate the elastance of the 
field in two ways: first, by adding the cells in parallel and the 
resultant laminae in series; secondly, by adding the cells in series 
and the resultant tubes in parallel. The first result is lower than 
the second. Readjust the positions of the lines of force and the 
equipotential surfaces until the two results are sufficiently close 
to one another; an average of the last two results gives very 
nearly the true elastance of the field. 

One difficulty in actually following out the foregoing method 
is that the changes in the assumed directions of the "field, that will 
give the best result, are not always obvious. Dr. Th. Lehmann 
has introduced an improvement which greatly facilitates the lay- 
ing out of a field. 1 While he has developed his method for the 
magnetic field, it is also directly applicable to the electrostatic 
field. We shall explain this method as applied to a two-dimen- 
sional field, though theoretically it is applicable to three-dimen- 
sional problems also. According to Lehmann, lines of force and 
equipotential surfaces are drawn at such distances that they 
inclose cells of equal elastance. Consider a slice, or a cell, in a 
two-dimensional field, o centimeters thick in the third dimension, 
and of such a form that the average length I of the cell in the 
direction of the lines of force is equal to its average width w in the 
perpendicular direction. The elastance of such a cell is always 
equal to unity, no matter whether the cell itself is large or small. 
This follows from the fundamental formula for elastance, which 
in this case becomes S = al/(a X w) = 1. 

The judgment of the eye helps to arrange cells of widths 
equal to their lengths, in proper positions with respect to each 

1 "Graphische Methode zur Bestimmung des Kraftlinienverlaufes in der 
Luft," Elektrotechnische Zeitschrift, Vol. 30 (1909), p. 995. 



Chap. XV] THE DIELECTRIC CIRCUIT 163 

other and to the electrodes; the next approximation is apparent 
from the diagram, by observing the lack of equality in the average 
width and length of the cells. Lord Rayleigh's condition is 
secured automatically, since the combination of cells of equal 
elastance leads to the same result, whether they are combined 
first in parallel or in series. After a few trials the space is properly 
ruled, and it simply remains to count the number of cells in series 
and in parallel. Dr. Lehmann shows a few applications of his 
method to practical cases of electrical machinery, and the reader 
is referred to the original article for further details. 

In a few simple cases, as for instance in determining the elas- 
tance between two parallel metallic cylinders of circular cross- 
section, or between two spheres, the principle of superposition of 
electric systems in equilibrium can be used, and the result obtained 
without trials. This principle is used in the determination of 
the capacity of transmission lines and cables, in the next two 
chapters. In two-dimensional problems, that is, in determining 
the shape of a field between two infinite parallel cylinders of any 
cross-sections whatever, the properties of conjugate functions can 
also be used in some simple cases; for further details see the 
references in Art. 10 above. 

Prob. 1. Sketch empirically the field between two infinite parallel 
cylinders of equal circular cross-section, the distance between the centers 
being a few times larger than the diameter. Determine the lower and 
upper limits of permittance per unit of axial length, and compare the 
results with the theoretical formula (320) given in Art. 63 below. 

Prob. 2. The terminal of a high-tension transformer consists of a 
long vertical rod connected to the winding, and a torus ring concentric 
with it, connected to the grounded case. The ring is of circular cross- 
section, and is placed near the center of the rod. Assuming the insula- 
tion in the whole field to be of the same permittivity, calculate by trials 
the elastance of the combination, with certain assumed dimensions of 
the rod and the ring. 

55. The Law of Flux Refraction. When an electrostatic 
flux passes from one dielectric into another of a different permit- 
tivity (Fig. 9, Art. 11), the lines of force suddenly change their 
direction at the dividing surface AB between the media, and 
in so doing they obey the law of refraction, which is 

tan 0i/tan 6 2 = /ci//c 2 (285) 

Here 6\ and 2 are the angles of incidence and refraction respec- 
tively, while ki and k 2 are the permittivities (relative or absolute) 



164 THE ELECTRIC CIRCUIT [Art. 56 

of the two media. A similar law is proved in Art. 11 above for 
the electric conducting circuit, and in Art. 41a of the Magnetic 
Circuit for the magnetic flux. The proof in the case of electro- 
static flux is similar in all respects to that given in Art. 11, if the 
student will use the words flux and flux density in place of current 
and current density, and permittivity in place of conductivity. 

Equation (285) shows that the lower the permittivity of a 
dielectric the more nearly do the lines of force in it approach the 
direction of the normal N1N2 at the dividing surface. In this 
way the path of a displacement between two given points is 
shortened in the medium of lower, and lengthened in that of 
higher permittivity, by such an amount in each case that the 
total permittance of the composite condenser is larger with 
refraction than without it. Hence, the existence of refraction is 
a necessary consequence of the general law of least resistance, 
mentioned in the preceding article. 

When mapping out an electrostatic field in two or more 
media, for instance, partly in a solid insulating material, partly 
in oil, and partly in air, the lines of force must be drawn so as to 
satisfy eq. (285) at the dividing surfaces. The permittance of 
the part of the circuit in any one of the media will not be a 
maximum, although the permittance of the whole combination 
must be a maximum. It will thus be seen that the problem, 
while quite simple in theory, is by no means an easy one in numer- 
ical applications, especially with the shapes of surfaces used in 
the construction of commercial high-tension apparatus. It is 
advisable for the student to train his eye in sketching lines of 
force in adjoining media of different permittivities, conforming 
the field at least approximately to eq. (285). This can be con- 
veniently done on available drawings of high-tension transformers, 
switches, lightning arresters, etc. 1 

66. The Dielectric Strength of Insulating Materials. The 
proportionality between stress and flux density, indicated, by 
eqs. (273) and (274), holds only up to a certain limit; in this 
respect it is similar to the proportionality between stresses and 
strains in an elastic body. After a certain limit of dielectric 
flux density or of voltage gradient has been exceeded, the material 

1 See also some interesting sketches and experiments in Professor W. S. 
Franklin's article on "Dielectric Stresses from the Mechanical Point of View," 
in the General Electric Review, Vol. 14, June, 1911. 



Chap. XV] 



THE DIELECTRIC CIRCUIT 



165 



weakens and finally breaks down. The phenomena of failure of 
electric insulation and the subsequent disruptive discharge are 
too well known to need a description here. 

The values of the critical voltage gradient G max and of the 
corresponding flux density D max , at which some of the more im- 
portant materials break down, are given in the last two columns 
of the table below. In designing insulation, the stresses must 
be kept well below these critical values, the factor of safety de- 
pending upon the importance of the apparatus, possibility of over- 
potentials, and the gradual deterioration of the insulation by heat, 
chemical action, moisture, and so forth. The values in the table 
are principally intended to give the student an idea of the order 
of magnitude of G ma x and D max . More accurate data will be found 
in electrical handbooks and pocketbooks; in important cases these 
design constants should be based upon test data obtained on the 
material in hand. 



Substance. 


Relative permit- 
tivity, or specific 
inductive capac- 
ity K. 


Rupturing volt- 
age gradient, 
G max, in kv - 
per mm. 


Rupturing values of 
dielectric flux den- 
si^' D maz, ^ mc - 
per sq. cm. 


Air 


1 

3-8 

5-8 

4.4 

2.2 

2.7 

2-2.2 

0.99 


3 

9-11 
24-40 
13-22 
28-35 
21-28 

14 


0.00265 
0.024 -0.078 
0.110 -0.280 
0.050 -0.085 
0.055 -0.068 
0.050 -0.067 
0.025 -0.027 


Glass, different kinds.. . 


Mica, natural and built up.. . . 

Porcelain 

Rubber, pure 


Rubber, vulcanized.. . 


Transformer oil 

Vacuum 







It will be seen from the second column of the table that the 
permittivities of solid and liquid dielectrics are larger than that 
of air; in other words, they are more yielding to electric stress 
than the air. This does not mean, however, that they break 
down at a lower voltage gradient than the air. On the contrary, 
the third and fourth columns show that the dielectrics commonly 
used in electrical engineering are considerably stronger electrically 
than the air, in that they can stand several times the electric stress 
and displacement at which the air breaks down. 

There does not seem to be any relation between the values of 
elastivity and critical voltage gradient. One indicates the elec- 
trical elasticity of the material, the other its ultimate strength. 
They are analogous to the modulus of elasticity and the rupturing 
stress respectively in the mechanics of materials. Air, from an 



166 THE ELECTRIC CIRCUIT [Art. 57 

electrical point of view, may be compared to a material of great 
stiffness, but one which breaks at a comparatively small elonga- 
tion. On the contrary, mica may be likened to a material which 
is comparatively yielding, but can stand a very large elongation 
before it is ruptured; so that, in spite of a smaller elastivity, 
a much higher stress is required to rupture mica than air. The 
student is advised to make clear to himself these two separate 
properties of dielectrics. A rational design of high-tension 
insulation depends essentially upon a distinct understanding of 
them. 

Dielectric strength may be properly given as the critical flux 
density, D ma x, but for practical purposes it is more convenient to 
express it as the critical voltage gradient, G max , at which the 
dielectric is broken down. When a dielectric is used for insula- 
tion in the form of thin sheets having a comparatively large 
radius of curvature, the flux density, and, consequently, the volt- 
age gradient, are practically uniform throughout, so that G max = 
Gave- When, however, the layer of dielectric is thick as compared 
to its radius of curvature, as for instance in the insulation of 
high-tension machines, or when air or oil are tested between 
two spherical terminals, the use of the average voltage gradient 
Gave = E/l leads to wrong results. The only proper way in 
this case is to calculate the voltage gradient for the place where 
it is a maximum, and to see that it does not exceed the critical 
value determined from previous tests. A breakdown in one 
point of the dielectric results in an increase of gradient in others, 
and possibly in a complete failure. 

Prob. 1. Show how the values in the last column of the table are 
derived from those in the two preceding columns. 

Ans. D max = 0.08842 KG max X lO" 2 . 

Prob. 2. A certain material stood about 82 kv. in a layer 3.7 mm. 
thick. What voltage gradient can be allowed in this material at a factor 
of safety of 2? Ans. 11 kv. per mm. 

Prob. 3. Assuming the relative permittivity of the insulation in the 
preceding problem to be 2.5, what is the density of energy at which the 
material is broken down? 

Ans. 5.45 X 10 -3 joules per cubic centimeter. 

57. The Electrostatic Corona. The phenomena which ac- 
company the electrical breaking-down of air deserve special 
mention in view of their great practical importance. When the 
voltage at the terminals of an air condenser is raised sufficiently 



Chap. XV] THE DIELECTRIC CIRCUIT 167 

high, a pale violet light appears at the edges, at the sharp points, 
and in general at the protruding parts having a comparatively 
small radius of curvature. This silent discharge into air, due to 
an excessive electrostatic flux density, is called the electrostatic 
corona. In the regions where the corona appears, the air is elec- 
trically " broken down" and ionized, so that it becomes a con- 
ductor of electricity. When the voltage is raised still higher the 
so-called brush discharge takes place, until the whole thickness 
of the dielectric is broken down, and a disruptive discharge, or 
spark, jumps from one electrode to the other. 

When the electrodes have projecting parts or sharp edges, the 
corona is formed at a voltage far below that at which the disrup- 
tive discharge occurs; the operating voltage of such devices is 
generally limited to that at which the corona forms. No corona 
is usually permissible in regular operation; first, because it may 
involve an appreciable loss of power; secondly, because the dis- 
charge, if allowed to play on some other insulation, will soon char 
and destroy it. There are cases, however, in which some corona 
formation is harmless. The air which is broken down becomes a 
part of the electrode, smoothes down the shape of the protruding 
metallic parts, increases their area, and thus reduces the danger- 
ous flux density and makes it more uniform. It is of advantage 
to operate certain parts of a very high-tension line at nearly the 
critical voltage. Any voltage rise on the line due to lightning 
or surges is automatically relieved by a corona loss into the 
atmosphere ; so that the line may be made self -protected, without 
lightning arresters. 

The formation of corona must be kept in mind in the design 
of high-tension insulation, and in high-potential tests. Shapes 
and combinations of parts which lead to high or non-uniform 
dielectric flux densities should be avoided. Fig. 48 shows the 
reason why the dielectric flux density, or the potential gradient, 
is higher near protruding parts. The equipotential surfaces, for 
obvious geometrical reasons, lie closer to each other near such 
parts, while at a reasonable distance from the electrodes the 
shape of the equipotential surfaces is not affected by small irregu- 
larities in the shape of the metallic parts. 

It will be seen from the table in the preceding article that the 
air is broken down when the voltage gradient exceeds 3000 volts 
per millimeter. Let this be the case at the point P (Fig. 48). 



168 THE ELECTRIC CIRCUIT [Art. 57 

The voltage gradient has this value only at the very surface of 
the conductor, because the lines of force immediately spread out 
in the air. Thus, only a very small portion of the air is broken 
down and becomes part of the conducting electrode. No visual 
corona is formed, however. Let now the voltage be raised still 
further; then the next layer of air is broken down and becomes 
part of the electrode. When a sufficiently thick layer of air is thus 
ionized, a visual corona is formed around the point P. Consider- 
ing the actual surface of the metal as the starting point, the volt- 
age gradient at that point now would seem to be higher than 3000 
volts per millimeter. This higher value is called the visual voltage 
gradient as distinguished from the disruptive voltage gradient of 
3000. The student should not be misled by these names. In 
reality the voltage gradient does not exceed 3000, because beyond 
this the air becomes part of the electrode; however, the concept of 
visual voltage gradient is convenient in calculations. 

In reality the phenomenon of ionization of air and formation 
of the corona is not as simple as described above, especially around 
conductors of small diameter, say less than 6 mm. The physical 
state of the layer of air adjacent to the conductor seems to be in 
some peculiar way affected by it, and the critical voltage gradient 
apparently depends in this case upon the diameter of the conductor. 
A discussion of numerical values and of physical theories is Out- 
side the scope of this book; and the student is referred for infor- 
mation to the numerous articles on the subject that appear in the 
leading periodicals, and in the transactions of the electrical engi- 
neering societies in this country and abroad. 1 

Quite extensive tests on corona formation, critical voltage, and 
the accompanying loss of power were performed by the General 
Electric Company, in 1910-11, and have been described by 
Mr. Peek. 2 The student is referred to his article for numerical 
data; the results are given on the first few pages of the article, 
and are illustrated by a numerical example. 

1 See, for instance, H. J. Ryan, " Open Atmosphere and Dry Transformer 
Oil as High-voltage Iusulators," in the Trans. Amer. Inst. Electr. Engrs., 
Vol. 30, Jan., 1911. This paper is a splendid exposition of the subject by 
one of the pioneer investigators of the corona, and contains numerous refer- 
ences to other articles on the subject. Professor J. B. Whitehead's experi- 
mental investigations are particularly noteworthy. 

2 F. W. Peek, Jr., " The Law of Corona and the Dielectric Strength of 
Air," Trans. Amer. Inst. Electr. Engrs., Vol. 30, July, 1911. 



Chap. XV] THE DIELECTRIC CIRCUIT 169 

Prob. 1. Assuming that under certain conditions a corona is formed 
when the dielectric flux density exceeds 0.0034 microcoulombs per square 
centimeter, calculate the factor of safety .of a 25-cycle transmission line 
for which the charging current is 0.12 amp. per kilometer, the diameter 
of the conductors being 12 mm. Solution: The line is charged during 
0.01 of a second, and the average charging current is 0.12/1.11 = 0.108 
amp.; hence, the maximum electrostatic displacement in the air is 1080 
microcoulombs per kilometer. The surface of each conductor is 377,000 
sq. cm. per kilometer, so that the density of displacement is 1080/377,000 
= 0.002865 microcoulombs per square centimeter, and the factor of 
safety is 34/28.65 = 1.20. 

58. Dielectric Hysteresis and Conductance. When an al- 
ternating voltage is applied at the terminals of a condenser, the 
dielectric is subjected to periodic stresses and displacements. 
If the material were perfectly elastic, no energy would be lost 
during one complete cycle, because the energy stored during the 
periods of increase in voltage would be given up to the circuit when 
the voltage decreased. In reality, the electric elasticity of solid 
and liquid dielectrics is not perfect, so that the applied voltage 
has to overcome some kind of molecular friction, in addition to 
the elastic forces. The work done against friction is converted 
into heat, and is lost, as far as the circuit is concerned. The 
phenomenon is similar to the familiar magnetic hysteresis, and is 
therefore called dielectric hysteresis. The energy lost per cycle 
is proportional to the square of the applied voltage, because both 
the displacement and the stress are proportional to the voltage. 

When stresses are well below the ultimate limit of the material, 
the loss of power caused by dielectric hysteresis is exceedingly 
small. Some investigators are even in doubt as to whether it 
exists at all. There is often an appreciable loss of power in com- 
merical condensers, but this loss can be mostly attributed to the 
fact that dielectrics are not perfect insulators. While their ohmic 
resistance is exceedingly high, as compared with that of metals, 
they nevertheless conduct some current, especially at high voltages. 
Thus, the observed loss of power and the heating of condensers 
may be simply ascribed to the PR loss in the insulation. More- 
over, small coronas can form at the edges and projecting parts, 
even at the operating voltage, and thus be an additional source 
of loss. Some small loss is also due to the ohmic resistance and 
eddy currents in the metallic sheets which compose the electrodes 
or plates of the condenser. 



170 THE ELECTRIC CIRCUIT [Art. 58 

An imperfect condenser, that is, one which shows a loss of 
power from one cause or another, can be replaced for purposes 
of calculation by a perfect condenser with an ohmic conductance 
shunted around it. This conductance, or " leakance," as some 
authors call it, is selected of such a value that the PR loss in it is 
equal to the loss of power from all causes in the given imperfect 
condenser. The actual current through the imperfect condenser 
is considered then as consisting of two components, — the leading 
reactive component through the ideal condenser, and the loss 
component, in phase with the voltage, through the shunted con- 
ductance. In this way, imperfect condensers can be treated graphi- 
cally or analytically, according to the ordinary laws of the electric 
circuit. 

Prob. 1. A certain kind of condenser shows a loss of power of about 
17.9 watts per microfarad, at 2200 volts, 25 cycles. By what fictitious 
conductance should an ideal condenser be shunted, in order to replace 
a condenser of this kind having a capacity of 1.5 mf.? 

Ans. 5.55 micromhos. 




CHAPTER XVI 

ELASTANCE AND PERMITTANCE OF SINGLE-PHASE 
CABLES AND TRANSMISSION LINES 

59. The Elastance of a Single-core Cable. A cross-section 
of a single-core cable is shown in Fig. 49. The round conductor 
in the center is assumed to be solid (not 
stranded) for the sake of simplicity. It is 
surrounded by a layer of insulation, and is 
protected on the outside by a lead sheath- 
ing. Let such a cable be subjected to a 
difference of potential between the core and 
the sheathing; for instance, let one pole of 
a battery be connected to the core and the 

other pole to the sheathing. Let it be re- 

• j x. a j xi. 'x.i. i-u i x. Fig. 49. A cross-section 

quired to rind the permittance or the elast- , . , 

n * of a single-core or con- 

ance of the dielectric for a certain axial ce ntric cable, 
length I of the cable. 

For reasons of symmetry, the lines of force are radial straight 
lines between the two metal surfaces, and the equipotential sur- 
faces are concentric cylinders. Consider the insulation to be sub- 
divided into concentric layers of infinitesimal thickness. The 
elastances of these layers are all in series, so that it is sufficient to 
express analytically the elastance of a layer having a radius x 
and thickness dx, and to integrate this expression between the 
limits a and b, where a is the radius of the core, and b is that of 
the inner surface of the sheathing. The elastance of the layer in 
question is 

dS = adx/(2wxl), (286) 

dx and 2 irxl being respectively the length and cross-section of the 
path of the radial flux. Integrating this expression between the 
limits a and b gives 

S= (a/2wl)Ln(b/a), (287) 

the abbreviation Ln standing for natural logarithm. 

171 



172 THE ELECTRIC CIRCUIT [Art. 59 

For practical calculations it is convenient to modify this formula 
in three respects; namely, (a) to introduce the relative permit- 
tivity K of the insulating material, (b) to express I in kilo- 
meters, and (c) to use common logarithms. Making these 
changes, we finally obtain 

S = C~ l = (41.45/20) log (b/a) megadarafs. . . (288) 

For the permittance (capacity) per kilometer we have accordingly 

C = C/l = 0.0241 K/\og (b/a) microfarads per kilometer. 1 (289) 

In some cases it is necessary to know the voltage across a 
certain part of the insulation, for instance between the radii r and 
r' . Applying formula (287) to this case, for I = 1 cm., we get 
S' rr = (a/2 t) Ln (r'/r) . The voltage drop E rr ' from r to r' is equal 
to this elastance multiplied by the electric displacement Q' per 
centimeter length of the cable. Or 

E rr > = &„' • Q' = W /2 7r) Ln (r'/r). . . (290) 

This formula finds its important application below in the calcula- 
tion of the permittance of single-phase and polyphase transmission 
lines. It is absolutely essential to agree in regard to the signs in 
eq. (290). In the applications that follow, Q' is taken with the 
plus sign when the positive displacement is directed from the con- 
ductor, and with the minus sign when it is directed towards the 
conductor. It is also important to write the distances r' and r in 
the order given, because interchanging r' and r in eq. (290) 
changes the sign of E rr \ 

If the insulation consists of two or more concentric layers of 
different materials, the elastances of the layers are calculated 
separately, according to formula (288), and then added in series. 
The permittance of the cable as a whole is the reciprocal of this 
resultant elastance. The same formulae apply to a concentric 
cable without sheathing, the outside conductor taking the place 
of the sheathing as far as stresses in the dielectric are concerned. 
With two cylindrical conductors side by side the elastance is cal- 
culated as shown in Art. 63 below. With three conductors the 
theory is rather difficult; as is also the case when the conduc- 
tors are not of circular cross-section. Those interested will find 

1 This simple derivation of the formula for the capacity of a single-core 
cable demonstrates in a particularly striking manner the usefulness of the 
concept of elastance. 



Chap. XVI] ELASTANCE OF CABLES AND LINES 173 

extensive literature on the subject in the European electrical 
magazines and proceedings of electrical societies. In practice, 
the permittance of such cables is usually determined by test. 

The distribution of the electric stresses in a single-core cable 
is of considerable practical importance. The total displacement 
Q being the same through every concentric layer of the dielectric, 
the flux density and consequently the stress is a maximum at the 
surface of the inner core. For a layer of radius x we have 

Q = D x -2irxl = const., (291) 

where D x is the density of displacement through that layer. 
Hence, 

D x x = const., (292) 

which means that the density of displacement is inversely pro- 
portional to the distance from the center. Since displacements 
are proportional to stresses (with a uniform insulation), we also 
have 

G x x = const (293) 

A useful relation between the total applied voltage E and the 
stress G x at a given point in the dielectric can be deduced from 
eq. (293). We have 

G x = const, /x; 

and if we multiply both sides by dx and integrate between a and 
b, remembering that voltage is the line integral of intensity, we 
obtain 

G x dx = E = (const.) Ln (b/a). 



f. 



Eliminating the constant between these two equations, gives 

G x = E/[x Ln (b/a)} (294) 

Equations (292) and (293) show that a homogeneous dielectric 
is fully utilized with regard to its dielectric strength only at the 
surface of the core, the stress gradually decreasing toward the 
periphery. * This condition could be helped by gradually increasing 
the elastivity of the material toward the sheathing, so as to in- 
crease the voltage drop and the stresses there. If the elastivity 
of each layer could be made exactly proportional to its radius, 
the stress G x would be the same throughout the dielectric. Such 
a condition would be an ideal one, with regard to economy in 



174 THE ELECTRIC CIRCUIT [Art. 59 

material, provided that the dielectric strength of the " variable 
insulation " were constant. 

This purely theoretical conclusion leads to the important 
practical question of the grading of insulation of cables. With 
high-tension cables, in which the thickness of insulation is large, 
it pays to provide two or more layers of different materials, utiliz- 
ing their permittivities and ultimate strengths in t'he most ad- 
vantageous manner. The problem is primarily to relieve] the 
stress near the inner core, and this is done by placing near it a 
layer of insulation of high permittivity, so as to cause a low volt- 
age drop. One case where the opposite arrangement would be 
advantageous is in a low-voltage cable in which it is desired to 
keep the total permittance as low as possible (for example, to 
obtain small capacity or low charging current at high frequen- 
cies). In this case the layer surrounding the core must have as 
high an elastance as possible, because it is this layer that contrib- 
utes most to the total elastance of the cable. With a clear 
understanding of these principles, the student will be able to 
design a graded insulation for given conditions, if he knows the 
properties of the available materials. 1 

Prob. 1. A single-core cable receives a charge of 1.18 millicoulombs 
per kilometer when a continuous voltage of 12 kv. is applied between 
the core and the sheathing. The core consists of a solid conductor the 
diameter of which is 5 mm.; the insulation is 9.5 mm. thick. Determine 
the value of the relative permittivity of the material of insulation, and 
the extreme values of the dielectric flux density. 

Ans. K = 2.78; B max = 0.00750; D min = 0.00156 microcoulombs 
per sq. cm. 

Prob. 2. The insulation used in the cable specified in the preceding 
problem breaks down at a flux density of 0.062 mc. per sq. cm. Show 
that the critical voltage for the cable is about 70 alternating kilo volts. 

Prob. 3. What is the ratio between the maximum and the average 
stress in the insulation in problem 1? Ans. 2.42. 2 

Prob. 4. Deduce formula (290) from the fact that the voltage is the 
line integral of the electric intensity. 

1 For a theoretical treatment of the grading of insulation, and for the 
bibliography of the subject, see H. S. Osborne, Potential Stresses in Die- 
lectrics (1910), a thesis presented to the Massachusetts Institute of Tech- 
nology for the degree of Doctor of Engineering. 

2 There is a tendency in practice to deal with average stresses even when 
the field is far from being uniform. The answer to this problem shows that 
one has to be careful in using an average value, unless its ratio to the maxi- 
mum stress is known. 



Chap. XVI] ELASTANCE OF CABLES AND LINES 175 

Prob. 6. Show by actual calculation that in the foregoing cable the 
maximum stress in the dielectric is reduced by increasing the diameter of 
the conductor to 7.5 mm., with the same diameter of the sheathing. 
This is in spite of the fact that the insulation becomes thinner, and 
consequently the average stress greater. 

Prob. 6. Referring to the preceding problem, show that it is of ad- 
vantage to make the ratio b/a about equal to e, where e = 2.71828 . . . 
is the base of the natural system of logarithms. If the diameter of the 
conductor be further increased, so that the ratio b/a becomes less than 
c, the maximum stress does not continue to decrease, but increases in- 
stead. Solution: The stress at the core is G a = E/[a Ln (b/a)] according 
to eq. (294). As a varies, G a reaches its maximum when dG a /da = 0. 
Differentiating, we get 

dGa/da = E [1 - Ln (b/a)] /[a Ln (b/a)]' = 0; 

whence, 1 — Ln (b/a) = 0, or b/a = e. 

Prob. 7. Explain the following deduction from the theorem stated in 
the preceding problem. In a concentric cable subjected to an excessive 
voltage, if the insulation is quite thick, the layer around the inner core is 
first gradually destroyed or charred up to a certain thickness, and then 
the rest of the insulation suddenly breaks down. With a thin layer of 
insulation no such phenomenon is observed. 

Prob. 8. A cable is provided with several concentric layers of insula- 
tion, the external radii of which are bi, b 2 , etc., and the relative permit- 
tivities, K h K 2 , etc. Show that the elastance of the cable is expressed 
by the formula 

S = (41.45/0 [Kr 1 log (h/a) + i^log (&2/&1) + i^log (b s /b 2 ) + etc.]. 

Prob. 9. Show that in a single-core cable the density of energy stored 
in the dielectric varies inversely as the square of the distance from the 
center. 

Prob. 10. A conductor 2 a cm. in diameter is surrounded by a con- 
centric metal cylinder of 26 cm. inside diameter. What alternating volt- 
age can be allowed between the cylinder and the conductor at a factor of 
safety k against the formation of corona? 

Ans. E = 18.4a (D c X 10 3 /k) log (b/a) effective kilovolts, where D c 
is the flux density in microcoulombs per sq. cm., at which corona is 
formed. 

Prob. 11. Show that the elastance of the dielectric between two con- 
centric spheres of radii a and b is equal to (<r a /4:irK) (1/a — 1/b) mega- 
darafs. 

Prob. 12. Show that with two concentric spheres the equation corre- 
sponding to (294) is G x = E/[x 2 (or 1 - 6" 1 )]. 

Prob. 13. Apply the formulae given in the text above to the theory of 
a condenser-type terminal. 1 

1 See A. B. Reynders, "Condenser Type of Insulation for High-tension 
Terminals," Trans. Amer. Inst. Electr, Engrs., Vol. 28 (1909), p. 209. 



176 THE ELECTRIC CIRCUIT [Art. 60 

60. The Elastance of a Single-phase Line. The general 
character of the electrostatic field between two infinite parallel 
conductors is shown in Fig. 50. The lines of force are arcs of 
circles extending from one metal surface to the other; the equi- 
potential surfaces are circular cylinders eccentric with respect to 
the conductors (see Art. 62 below). It is required to calculate 
the elastance of the air between the two conductors, for a unit 
axial length of the line. Knowing this elastance, the charging 
current of the line can be calculated for a given frequency. This 
elastance, or its reciprocal, the permittance, is used in the pre- 
determination of the regulation of a transmission line (Art. 68 
below). 

We shall consider in this article the usual practical case in 
which the radius a of the conductors is small as compared with 
the interaxial distance b. It is shown in Art. 63 below how to 
determine the elastance when the diameters of the cylinders are 
comparatively large. 

For purposes of analysis it is convenient to consider the field 
shown in Fig. 50 as the result of the superposition of two simple 
radial fields similar to that in Fig. 49. Consider the conductor 
A, together with a concentric cylinder of an infinitely large radius, 
as one electric system. Let the conductor B with a similar con- 
centric cylinder form another independent system. Let the con- 
ductor A be connected to the positive pole of a battery of voltage 
E, the conductor B to the negative pole, and the two cylinders at 
infinity to the middle point of the battery. In the first concentric 
condenser the displacement of positive electricity is from the con- 
ductor A to the infinite cylinder, while in the second system the 
positive displacement is from the infinite cylinder toward the con- 
ductor B. The displacements due to the two systems are equal 
and opposite at the two infinite cylinders, and the cylinders them- 
selves coincide at infinity, because the distance AB between their 
axes is infinitely small as compared with their radii. Hence, the two 
displacements at the cylinders cancel each other, and the combina- 
tion of the two cylindrical condensers is electrically identical with 
the two given parallel conductors A and B. 

In a medium of constant permittivity the resultant stress or 
voltage gradient, produced at a point by the combined action of 
two or more independent electric systems, is equal to the geometric 
sum of the stresses produced at the same point by each system. 



Chap. XVI] ELASTANCE OF CABLES AND LINES 



177 



This principle of superposition can be considered either as an ex- 
perimental fact or as an immediate consequence of the fact that 
in a medium of constant permittivity the effects are proportional 
to the causes. This principle being true for electric intensities, 
the component flux densities at a point are also combined accord- 
ing to the parallelogram law, because they are proportional to 
the intensities. Hence, the resultant electrostatic flux can be re- 
garded as the result of the superposition of the fluxes created by 




Fig. 50. The electrostatic field produced by a single-phase trans- 
mission line. 

the component systems. Furthermore, the actual voltage be- 
tween any two points in the dielectric is the algebraic sum of the 
voltages due to the component systems, because each voltage is 
the line integral of the corresponding voltage gradient, and the 
principle of superposition is valid for these gradients. This line 
integral is a function only of the positions of the two points, and 
is independent of the path along which the integration is per- 
formed. This latter fact is very convenient in applications of the 
principle to the solution of problems. 



178 THE ELECTRIC CIRCUIT [Art. 60 

In order to be able to apply the formulae deduced in the pre- 
ceding article, it is essential that the diameters of the two wires be 
small as compared with the distance between them. The reason is 
that each component system is supposed to possess a radial field, 
in spite of the presence of the other conductor. This is practically 
true when the second conductor is so small, or so far distant from 
the first, that the infinite permittivity of its material does not 
appreciably distort the radial field. To be more precise, the dis- 
tortion of the radial component field originating from each con- 
ductor, caused by the presence of the other, must be negligible. 

It is sufficient to calculate the elastance of that part of the 
system between one of the conductors and the neutral plane of 
symmetry 00' ', the total elastance being equal to twice that 
value. This we can do by computing the voltage needed to 
produce a displacement Q' per unit length of the line. The volt- 
age between the surface of the conductor A and the point -N in 
the plane of symmetry 00' is equal to \ E. On the other hand, 
the same voltage can be expressed as the sum of the voltages due 
to the two component systems. Referring to eq. (290), let the 
distance r' refer to the point N, and let r refer to a point on the 
surface of the conductor A. Then, as far as the first component 
system is concerned, the voltage between A and N is equal to 
(<jQ'/2 t) Ln (J b/a), where Q' is the actual displacement per unit 
length of the line. In the second system, the voltage between the 
same two points is — (aQ ' /2 if) Ln (| 6/6) . The minus sign is 
due to the fact that the displacement in the second system is 
toward the conductor B, and hence must be considered as negative 
if that at the first conductor is regarded as positive. The ratio 
r'/r for the second system is more accurately equal to \ 6/(6 — a), 
but a, being by supposition small as compared to 6, is neglected in 
the denominator. Equating the sum of the preceding two expres- 
sions for the voltage between A and N to the actual voltage J E, 

we get 

i E = (aQ'/2 t) Ln (b/a) (295) 

Hence, the elastance between one of the conductors and the 
neutral plane 00', for a unit of axial length, is 

S' = (CO" 1 = \ E/Q' = (cr/2 tt) Ln(6/a), . (296) 
or, with air as the dielectric, 

S' = (CO" 1 = 41.45 log (b/a) megadarafs per kilometer. (297) 



Chap. XVI] ELASTANCE OF CABLES AND LINES 179 

The corresponding permittance is 

C" = OS') -1 = 0.0241/log (b/a) microfarads per kilometer. (298) 

When using these formulae, one must not forget that the per- 
mittance is proportional to the length of the line, while the elas- 
tance varies inversely as the length of the line. The total elastance 
for a unit length between the two conductors is equal to 2 S', the 
corresponding permittance being J C . 

Prob. 1. For a few standard spacings and sizes of conductor, check 
the values of permittance given by eq. (298) with those tabulated in an 
electrical pocketbook. 

Prob*. 2. For some assumed values of a, b, and E, corresponding to 
an actual transmission line, plot a curve of values of the voltage gradient 
along the line AB, and also draw the horizontal straight line represent- 
ing the average gradient E/b. 

Hint: At a distance x from A the intensity due to the system A is 
aQ' I (2 wx); that due to the system B is <rQ'/ [2 ir (b — x)], both inten- 
sities being directed from left to right. 

Prob. 3. In Fig. 50 let A and B be small spheres, instead of cylinders. 
Show that the elastance between one of the spheres and the neutral 
plane 00' is equal to (o-/4 x) (1 /a — 1/6). Hint: Apply the principle of 
superposition, as in the text above, and utilize the solutions of problems 
11 and 12 of the preceding article. 

Prob. 4. In a transmission line the wire B is split into two separate 
conductors B x and B 2 , connected in parallel. The spacings A — Bi, 
A — B 2 , and Bi — B 2 are equal to b x , b 2 , and b 12 respectively. Show how 
to calculate the total permittance per unit length of the line, using the 
method of superposition. Solution : Let Q', Qi, and Q 2 be the displace- 
ments issuing per unit length of the conductors A, Bi, and B 2 respectively. 
Resolve the given system into three systems, with the three given con- 
ductors each concentric with a cylinder of infinite radius. Then we have 
the following three conditions: (a) Q' -f- Q/ + Q 2 = 0, because electricity 
behaves as an incompressible fluid; (b) the given voltage E between 
A and Bi is the sum of the partial voltages for the three component 
systems, each expressed according to eq. (290) ; and (c) the same is true 
for the voltage E between A and B 2 . 1 From these three equations the 
quantities Qi and Q 2 are eliminated, and the required elastance is de- 
termined from the resultant equation, as the ratio of E to Q'. 

Prob. 5. Show how to calculate the elastance between two small 
cylinders or spheres of unequal radii. 

Prob. 6. Analyze the formal mathematical reason for which the elec- 
trostatic equipotential lines in Figs. 49 and 50 coincide with the magnetic 
lines of force, and vice versa. Compare Figs. 46 and 47, Arts. 59 and 
60, in the Magnetic Circuit. 

1 Or else we may use as condition (c) the fact that the resultant voltage 
between B x and B 2 equals zero. 



180 THE ELECTRIC CIRCUIT [Art. 61 

61. 1 The Influence of the Ground upon the Elastance of a 
Single-phase Line. When the ground is used as the return con- 
ductor of a circuit, for instance in single-phase railways and in 
telegraph lines, the elastance of the circuit is calculated by assum- 
ing the ground to be a good conductor of electricity; in other 
words, its permittivity is assumed to be infinitely large. This 
gives a larger permittance than any other assumption, and con- 
sequently a value which is on the safe side. According to the 
law of refraction (Art. 55) the lines of force from the metallic con- 
ductor enter the ground at right angles to its surface; so that the 
field has the shape shown in Fig. 50, between one of the wires 
and the plane of symmetry 00', which in this case represents the 
surface of the ground. This leads to Lord Kelvin's method of elec- 
tric images, which we shall use in its simplest form only. 

When it is required to find the shape of the field, or the elastance 
between a conductor such as A and an infinite conducting surface 
such as 00', first locate a fictitious conductor B, which is the elec- 
tric image of the conductor A; that is, B is located as if it were 
the optical image of A in the plane mirror 00' . Furthermore, if 
A has a potential of E volts above that of 00', take the potential 
of B as E volts below that of 00' , the voltage between A and B 
thus being 2 E volts. Having located B, the conducting plane 
00' is removed, and the field between A and B is determined. 
The part of the field between A and 00' has real existence, that 
between 00' and B is fictitious. 

The validity of this principle in the case under consideration 
becomes evident by the following reasoning: Let a voltage of 
2 E be maintained between A and B by means of a battery. 
Place an infinite conducting sheet of negligible thickness so as 
to coincide with the equipotential plane 00'. The field is not 
affected thereby, the lines of displacement being normal to this 
sheet. Connecting the sheet to the middle point of the battery 
does not in any way disturb the field. Now the field between A 
and the sheet 00' is maintained by one half of the battery, that 
between B and 00' by the other half. Both halves are in equi- 
librium independently of each other, so that the conductor B with 
its half of the battery may be removed without disturbing the 
field between A and 00'. Conversely, to find the field between 

1 The rest of this chapter may be omitted if so desired, as it is not neces- 
sary to an understanding of the remainder of the book. 



Chap. XVI] ELASTANCE OF CABLES AND LINES 181 

A and a conducting sheet 00', the latter is replaced by a fictitious 
conductor B, so as to reduce the conditions to those investigated 
in the previous article. For a general discussion of the principle 
of electric images, see any standard work on the mathematical 
theory of electricity and magnetism. 

Applying this principle to the case of a single-phase line with 
ground return, we see immediately that all of the formulae de- 
duced in the preceding article hold true, provided that we put 
b = 2 h, where h is the elevation of the conductor above the 
ground. Since this elevation is usually quite considerable, it will 
be seen that the elastance of the circuit is larger, and the charging 
current smaller as compared to the case of a metallic circuit 
having a comparatively small spacing. 

The next case to be considered is that of the elastance of a 
metallic return line, as reduced by the proximity of the earth 
(Fig. 51). The elastance is reduced as compared to that in Fig. 
50 because part of the medium of finite elastivity (air) is replaced 
by the ground, which is assumed to be of zero elastivity, or a 
good conductor of electricity. It will be seen from the figure that 
the lines of force are deflected toward the ground, where they 
find a path of less elastance. 

The total elastance between the conductors A and B is cal- 
culated, using again the method of electric images. The lines of 
force meet the ground at right angles, and its surface is one of 
equal potential. The field above the ground would be the same 
if the ground were removed and replaced by the electric images 
A' and B' of the wires, the polarity of the images being that indi- 
cated in the sketch. The surface of the ground becomes now 
a plane of symmetry. The fictitious field below the ground is 
indicated by the dotted lines. The dielectric field may now be 
considered as if due to a superposition of four systems, each consist- 
ing of one of the conductors and a cylinder at infinity. Applying 
eq. (290) for each of these four systems, we find that the voltage 
between A and B, 

due to system A, is + (aQ'/2 it) Ln (b/a); 
due to system B, is — (ctQ'/2 t) Ln (a/b) ; 
due to system A', is - {<rQ' /2 ir) Ln (2 d/2 h) ; 
due to system B', is +(<rQ'/2 tt) Ln (2 h 2 /2 d) ; 

where 2d = AB' = A'B. The actual voltage between A and B 



182 



THE ELECTRIC CIRCUIT 



[Art. 61 



being equal to E, we have, by adding the preceding four ex- 
pressions, 

E = (aQ'/2 tt) [2 Ln (b/a) - Ln (d 2 /hh 2 )]. . . (299) 
One half of the elastance between A and B is 

& = i W = 0/2 tt) [Ln (b/a) - J Ln (tf/hfo)]. (300) 




/■■ / \ 



Fig. 51. The electrostatic field due to a single-phase line AB, as affected 
by the proximity of the ground. A' and B' are the electric images of A and B. 

This expression is identical with formula (296), except for the last 
term, which represents the reduction in elastance due to the pres- 
ence of the ground. When the distances h and h 2 to the ground are 
large as compared to the spacing b, the ratio d 2 /hji 2 differs but 
little from unity, and the correction is small because the logarithm 



Chap. XVI] ELASTANCE OF CABLES AND LINES 183 

of unity is equal to zero. Equation (300) can be written also in 
a simpler form by combining the two logarithms into one. We 
obtain then 

S' = (a/2 7r)Ln(b c /a), (301) 

where b c stands for the corrected spacing, determined from the ex- 
pression 

b c = b (y/kyhi/d) (302) 

We have thus arrived at the following simple rule: The elastance 
and permittance of a single-phase line, with the effect of the ground 
considered, are expressed by the same formulce (296) to (298) as 
though this effect were ignored, provided that the actual spacing b is 
replaced by the corrected spacing b c given by formula (302) or (305). 
In practice, the values of hi, h 2 , and b are known, and it is de- 
sirable to avoid the use of the quantity d in the foregoing formula. 
Applying a well-known theorem of elementary geometry, we have 
from the triangle AA'B 



A'B* = AA ,Z + AB l + 2 AA' X AP, 
or 

4 c/ 2 = 4hi 2 + b 2 + 4:hi(h 2 - hi), 

from which 

4 d 2 = b 2 + 4 foil*. 
Hence 

d 2 /foh 2 = 1 + \ b 2 /hh 2 (303) 

Equation (300) becomes then 

S' = (a/2 tt) [Ln (b/a) - J Ln (1 + J b 2 /hh 2 )], . (304) 

and from eq. (302) we have 

b c = b/Vl + | b 2 /hih 2 . (305) 

When tables of capacity for standard spacings are used, as tabu- 
lated in various reference books, the correction for the influence 
of the ground will be found convenient in the form shown in 
problem 2 below. 

Prob. 1. For various usual spacings and sizes of conductor, calculate 
the per cent error introduced in computing the permittance of a trans- 
mission line by neglecting the influence of the earth in the most unfavor- 
able cases. Select the conductors either in a vertical or in a horizontal 
plane, whichever arrangement in your opinion is more affected by the 
ground. 



184 THE ELECTRIC CIRCUIT [Art. 62 

Prob. 2. When permittances are taken from standard tables, it is not 
convenient to use the corrected spacing b c , because capacities are tabu- 
lated for standard spacings only. In this case it is convenient to repre- 
sent the elastance given by eq. (304) in the form S c ' = S' — s, where S' 
is the reciprocal of the value of capacity found in the tables, and s is the 
correction due to the presence of the ground. Deduce a simple form of 
this correction, when it is small. Solution : Expanding the natural loga- 
rithm according to the series Ln (1 + x) = x — % x 2 + J x z — etc., we find 
that the corrections = 9 Ln (1 + \ b 2 /hM = 9 [J b 2 /hh 2 - §(} b 2 /hji 2 ) 2 
+ etc.] in megadarafs per kilometer of one conductor. S' must of course 
be taken also in megadarafs per kilometer of one conductor. 

Prob. 3. Deduce formulae for the influence of the ground in the case 
of small spheres in place of the cylinders. 

62. The Equations of the Electrostatic Lines of Force and 
Equipotential Surfaces Produced by a Single-phase Line. In 

Fig. 50, let P be a point on the line of force AN'PB, of which 
we desire to find the equation. Let us calculate the total flux 
which passes from conductor A to B between the plane of sym- 
metry AB, and the surface of force on which the point P is 
located. 1 

Let the axial length for which the flux is determined be equal 
to one centimeter. This flux may be considered as the resultant 
of the fluxes due to the systems A and B. The radial flux pass- 
ing between AB and P, due to the component system having 
the center A, is equal to Q'di/2ir, and is directed from left to 
right. The flux due to the B system is equal to Q'd 2 /2 ir, and is 
also directed from left to right, B being the negative conductor. 
The total or the actual flux between the surfaces of force AB and 
AN'PB equals (Q'/2 r) (0i + 2 ). Since this flux does not depend 
upon the position of the point P, provided that the point is taken 
upon the line of force under consideration, we have 

0i + 02 = const (306) 

for all points on a line of force. This is the equation of the line 
of force. For different lines of force, the value of the constant is 
different. In the triangle APB, the angle o> is supplementary to 

1 It is convenient to speak of a surface formed by lines of force as a sur- 
face of force. For instance, a line of force such as AN'PB, should it move in 
a direction perpendicular to the plane of the paper, would form a cylindrical 
surface, which we shall call a surface of force, by analogy with a line of force. 
On the other hand, we shall call a line such as CPC an equipotential line, to 
distinguish it from the corresponding equipotential surface which it repre- 
sents in the sketch. 



Chap. XVI] ELASTANCE OF CABLES AND LINES 185 

the sum of the angles 0i and 2 , so that condition (306) may also 
be written 

co = const (307) 

This represents the arc of a circle passing through A and B, of 
which co is the inscribed angle. It is thus proved that the lines of 
force are arcs of circles passing through A and B. 

For points on the line of symmetry 00' the angles 0i and 2 are 
equal, so that the total flux corresponding to a certain angle 0i 
is (Q'/2 7r) (2 0i) = Q'Oi/t. This fact permits us to mark on the 
line 00' the intersections of the surfaces of force between which 
are included definite fractions of the total flux Q'. For instance, 
let it be desired to draw a line of force such that the flux between 
it and the plane AB shall equal one sixth of the total flux. One 
sixth of 180 degrees are 30 degrees; we therefore draw from A a 
straight line at an angle of 30 degrees to AB, and through its inter- 
section with 00' draw an arc of a circle passing through A and B. 
In this way, the total flux," or what is the same, the total per- 
mittance between A and B, can be divided into any number of 
equal or unequal permittances in parallel. 

To prove that the equipotential lines are also circles, take 
again a point P determined by the distances n and r 2 from A and 
B respectively. If the point C lies on the same equipotential line, 
the voltage between P and C is equal to zero, so that, applying 
eq. (290) for the two component systems, we get 

(oQ'/2t) Ln (n/AC) - (aQ'/2ir) Ln (r 2 /BC) = 0, 

from which 

Ln (n/AC) = Ln (r 2 /BC), 
or 

n/rj = AC/BC = const (308) 

This is the equation of an equipotential line in " bipolar " co- 
ordinates; the curve is such that the ratio of n to r 2 remains 
constant. This constant is different for each equipotential line, 
because each line has its own point C. 

Equation (308) may be proved to represent a circle, by select- 
ing an origin, say at A, and substituting for T\ and r 2 their values 
in terms of the rectangular coordinates x and y. The following 
proof by elementary geometry leads to the same result. Produce 
AP and lay off PD = PB = r 2 . According to eq. (308), BD is 
parallel to CP, and consequently PC bisects the angle APB =■ co. 



186 THE ELECTRIC CIRCUIT [Art. 62 

Let the point C lie on the same equipotential line with C; then 
the voltage between P and C is also equal to zero, and by analogy 
with eq. (308) we have 

n/r 2 = AC'/BC = const. .... (309) 

By plotting PD' = r 2 (not shown in the figure) along PA, in the 
opposite direction from PD, and connecting T>' to B, one can show 
as before that PC bisects the angle BPD = 180° - a, But the 
bisectors of two supplementary angles are perpendicular to each 
other; consequently, CPC is a right angle, and the point P lies 
on a semicircle drawn on the diameter CC. This semicircle is 
the equipotential line itself, because all points, such as P, which 
are determined by C and C ', must lie on it. The semicircle below 
A B evidently belongs to the same equipotential line. 

From eqs. (308) and (309) the following expressions are ob- 
tained for the radius R of the equipotential line under considera- 
tion. 

R = i - (Bc/Acy (310) 

or 

R = i + bc/ac: (311) 

so that the equipotential line can be easily drawn for a given 
C or C. 

Let it be required to calculate the elastance of the slice of die- 
lectric between the neutral plane 00' and the equipotential sur- 
face passing through a given point C. It is sufficient to find the 
expression for a unit axial length, knowing that the elastance is 
inversely proportional to the length of the conductors. 

Write the expression for the voltage between the points N and 
C, using again eq. (290). For the systems A and B we have 

E NC = (aQ'/2ir) [Ln (AC /AN) - Ln (BC/BN)], 

or, since AN = BN, 

E N c = (aQ' /2w)Ln (A C/BC). . . . . (312) 

From this equation we see that the elastance per centimeter 

Snc = E NC /Q f = (a/2 tt) Ln (AC/BC). . . (313) 

From this expression, the elastance between any two equipotential 
surfaces can be calculated, by computing first the elastance be- 



Chap. XVI] ELASTANCE OF CABLES AND LINES 187 

tween each of the surfaces and the plane of symmetry 00', and 
then taking either the sum or the difference of these elastances, 
depending upon the positions of the two given surfaces; that is, 
whether they lie on different sides or on the same side of the 
plane 00'. 

It has been explained above how to divide the field by surfaces 
of force into permittances of desired values, these permittances be- 
ing proportional to the angles 0i or 6 2 , determined by the point N' 
on the neutral plane. Knowing now how to subdivide the field 
into elastances of desired values by equipotential surfaces, the 
student can without difficulty calculate the permittance or the 
elastance of a given slice in the field between two equipotential 
surfaces and two surfaces of force. 

Prob. 1. For an assumed size of the conductors and a spacing used in 
extra-high-tension transmission lines, draw a set of lines of force and equi- 
potential lines (Fig. 50) such as to divide the total voltage and the total 
electrostatic flux into 10 equal parts. 

Prob. 2. Let A and B (Fig. 50) be two very small wires at a distance 
of 90 cm. between centers. What is the permittance of the slice NN'PC 
if NN' = 25 cm.; NC = 32 cm.; and the axial length is 180 m.? 

Ans. 0.000861 mf. 

Prob. 3. Show that the lines of force between two small spheres are 
not circles, but curves the equation of which is cos Q x + cos 2 = const. 

Prob. 4. Show that the equipotential surfaces in the case of two 
small spheres are represented by the equation 1/n — l/r 2 = const. 

Prob. 5. Show how to draw in a given case the field shown in Fig. 51. 
Solution: Draw a set of n lines of force due to the system AB alone, the 
same as in Fig. 50. Let the flux Q' be divided into n equal parts, so that 
the flux between the adjacent surfaces of force is Q' '/n. Draw a similar 
set of lines of force for the system A'B' . The equation of a line of force 
in the system AB is a = const.; that in the system A'B' is &/ = const. 
According to the principle of superposition, the equation of a line of force 
in the resultant field is co — J = const., the minus sign being due to the 
fact that A' is negative if A is positive. Let the point of intersection of 
two lines of force a = C and co' = C be the starting point for draw- 
ing a line of force in the resultant field. Then the point of intersection 
of the next lines co = C + ir/n and J = C -+- ir/n also belongs to the 
same line of force in the resultant field, because for both points co — J = 
C — C. In other words, the lines of force in the resultant field are 
diagonal curves with respect to the lines of force in the component fields, 
and may be drawn from intersection to intersection. A similar construc- 
tion holds for equipotential surfaces. The student is strongly urged to 
try this construction for some assumed data, because the method of 
diagonal curves is generally applicable when a given field can be resolved 
into two simpler fields. 



188 THE ELECTRIC CIRCUIT [Art. 63 

63. The Elastance between Two Large Parallel Circular 
Cylinders. The formulae derived in Art. 60, for the elastance 
and permittance of a homogeneous medium between two parallel 
cylinders, hold true only when the diameters of the cylinders are 
small as compared to the interaxial distance, for the reason there 
explained. When the diameters of the cylinders are compara- 
tively large, the elastance is derived by reducing the conditions to 
those obtaining in Art. 60. 

Let A and B (Fig. 50) represent as before two conductors of 
very small diameter, and let a difference of potential of 100 volts 
be maintained between them by means of a battery. Let the volt- 
age between the conductor B and the equipotential surface CPC 
be 20 volts. Place an infinitely thin metal sheet so as to coincide 
with this surface, and connect this sheet to a point of the battery 
such that the voltage between it and the conductor B still remains 
equal to 20 volts. These changes do not affect the electrostatic 
field either inside or outside the surface CPC' y the displacement 
being normal to this surface. Now remove the conductor B al- 
together, leaving a difference of potential of 80 volts maintained 
by the battery between the conductor A and the cylinder CPC '. 
The field outside the cylinder is not affected; that inside of it has 
entirely disappeared. We have now a field between the cylinder 
A of very small diameter and the cylinder CPC of a compara- 
tively large diameter. Take now another equipotential surface, 
for instance KMK' ', symmetrical with CPC, place a metal, cylinder 
so as to coincide with it, and connect it to a tap on the battery, 
so that the same difference of potential of 20 volts remains be- 
tween this cylinder and the conductor A . The field is not altered 
by this connection, and now the conductor A may be removed. 
Thus, we finally obtain a field between two cylinders of compara- 
tively large diameter. The difference of potential between the 
cylinders is only 60 volts, while the original difference of potential 
between the conductors A and B was 100 volts. 

Conversely, let the cylinders CPC and KMK' be given, and 
let it be required to find the shape of the field between them, and 
the elastance of this field. The problem is reduced to that of 
finding the positions of the infinitely small eccentric conductors 
A and J5, with respect to which the given cylinders are equi- 
potential surfaces. Then the field is mapped out according to 
the formulae given in the preceding article, leaving out the space 



Chap. XVI] ELASTANCE OF CABLES AND LINES 189 

inside the cylinders. The elastance between one of the large 
cylinders and the plane 00' is calculated by using formula (313). 
This method is applicable whether the two cylinders are of the 
same radius or not, and whether one is outside or inside of the 
other. It is always possible to find the positions of the lines A 
and B with respect to which the given cylinders represent equi- 
potential surfaces. The details of the calculation are given below. 
Consider first the case of two cylinders CPC and KMK' of 
the same diameter d; let the distance between the centers p and 
q of these cylinders be equal to c. In order to use eq. (313), it is 
necessary to express AC and BC through the given quantities c 
and d. According to eqs. (308) and (309), we have 

AC/BC = AC/BC (314) 

All the quantities which enter into this equation can be expressed 
through one unknown length, for instance BC. We put 

BC = AK = x; 

then 

AC = CK + AK = (c - d) + x\ 

AC = x + c; \ • ■ • (315) 

BC = d- x. 

Substituting these values into eq. (314), and solving the resulting 
quadratic equation for x, we obtain, retaining the positive value 

only, 

BC = AK = x = \ [- (c -d) + Vc 2 - d 2 ] 

= id[-(a-l) + V a 2 -l], (316) 

where the ratio of the interaxial distance to the diameter is de- 
noted by a, or 

a = c/d. ....... (317) 

By substituting this value of x into the expression for AC in eqs. 
(315), we find 



AC = i[(c - d) + Vc 2 - d*} = i d[(a - 1) + Va 2 - 1], (318) 
so that 



AC/BC = [(a - 1) + Va 2 - l]/[ - (a - 1) + ^a 2 - 1]. 

This expression can be simplified by multiplying both the numer- 
ator and the denominator by the value of the numerator, so 



190 THE ELECTRIC CIRCUIT [Art. 63 

as to get rid of the square root in the denominator. The 

result is 

AC/BC = a + W - 1 (319) 

The expression (313) for the elastance between one of the cylinders 
and the plane of symmetry, per unit of axial length, becomes 



S' = (a/2 tt) Ln [a + Va 2 - 1]. . . . (320) 

Those familiar with hyperbolic functions will notice that the pre- 
ceding equation can be simplified into 

S' = (a/2ir)Cosh- l a. ..... (321) 

Since tables of hyperbolic functions are readily available, the 
evaluation of elastance is simpler in this form than it is if eq. (320) 
is used. 1 

When the diameter of the conductors is small as compared to 
the interaxial distance, a is a large quantity, and unity under the 
radical sign in eq. (320) may be neglected. This equation be- 
comes then practically identical with eq. (296). For large values 
of a, the term (1 — l/a 2 )*, obtained by factoring in expression (320), 
is conveniently expanded according to the binomial theorem, the 
result being 

S' = (<7/2tt) Ln (2 a - J a~ l - § a" 3 - T \ or 5 - . . .). (322) 

With the exception of 2 a, all of the terms in parentheses are 
small corrections to the result. 

Let now the diameters of the two given cylinders be different. 
In addition to relation (314), we also have 

BK/AK = BK'/AK' (323) 

It is necessary in this case to introduce two unknown quantities, 
BC = x and AK = y. Equations (315) are modified accordingly. 
All of the quantities in eqs. (314) and (323) are expressed through 
x and y, and then these two equations are solved together for x 
and y. After this, the elastance between each cylinder and the 
plane 00' is expressed by using eq. (313). 

1 Dr. A. E. Kennelly, " The Linear Resistance between. Parallel Conduct- 
ing Cylinders in a Medium of Uniform Conductivity," Proceedings Amer. 
Philosophical Soc, Vol. 48 (1909), p. 142; also his article on "Graphic Rep- 
resentations of the Linear Electrostatic Capacity between Equal Parallel 
Wires," Electrical World, Vol. 56 (1910), p. 1000. See also his book on Ap- 
plications of Hyperbolic Functions to Electrical Engineering (1912). 



Chap. XVI] ELASTANCE OF CABLES AND LINES 191 

In some cases it is required to calculate the dielectric flux 
density at a point in the field between the cylinders, or at the sur- 
face of one of the cylinders. Let P (Fig. 52) be a point in the 
field between two parallel cylinders, small or large; the flux density 
at P is the geometric sum of the densities due to the systems A 
and B. The flux density due to the system A is 

while that due to B is 

D 2 = Q7(2?rr 2 ). 

These component densities are directed as shown in Fig. 52. 
The resultant density D is directed along the tangent to the line 




b B 

Fig. 52. Dielectric flux density at a point, determined by the method 
of superposition. 

of force through P. From the preceding two equations, we have 
the relation 

Di:D 2 = r 2 :r h 

so that the triangles APB and Pmn are similar. The correspond- 
ing sides are marked with one, two, and three short lines respec- 
tively. From these triangles we can write 

D:D 1 = b: r 2 , 

or, substituting the foregoing expression for D h 

D = Q , b/(2Trr l r 2 ) (324) 

From this expression, the flux density can be calculated at any 
point in the field or on the surface of one of the cylinders. Mul- 
tiplying the flux density by the elastivity of the medium, the cor- 
responding dielectric stress is obtained. It must be kept well in 
mind that b, n, and r 2 refer to the points A and B, and not to the 
centers p and q of the actual cylinders. 



192 THE ELECTRIC CIRCUIT [Art. 63 

Prob. 1. Take two equal cylinders at a comparatively short distance 
apart, and (a) calculate the permittance per meter of the axial length; 
(b) divide the field into 10 equal elastances in series and into 10 equal 
permittances in parallel; (c) plot a curve of the flux density distribu- 
tion on the surface of one of the cylinders. 

Prob. 2. Show that on an equipotential surface surrounding A, and 
consequently on the corresponding metal surface, the flux density varies 
inversely as n 2 . 

Prob. 3. Show how to calculate the permittance between a large cylin- 
der and a given infinite plane. 

Prob. 4. Show that A and B are inverse points with respect to any 
equipotential circle; this means that the radius qC is the geometric mean 
between the distances qB and qA, and the radius pK is the geometric 
mean between the distances pA and pB. This is true whether the radii 
qC and pK are equal or not. 

Prob. 5. Extend the theory given in this article to the calculation of 
the elastance and flux density distribution between two large spheres. 
Consult the chapters on electrostatics in some standard work on the 
mathematical theory of electricity and magnetism. 



CHAPTER XVII 

EQUIVALENT ELASTANCE AND CHARGING CURRENT 
OF THREE-PHASE LINES 

64. Three-phase Line with Symmetrical Spacing. Consider 
an unloaded three-phase line, and let the three conductors be 
denoted by A, B, and C respectively. There is a displacement of 
electricity between each pair of conductors, and since the three in- 
stantaneous voltages are different, the displacements between the 
three pairs of conductors at any instant are also different. The 
three sets of lines of force are relatively displaced and the flux 
density varies from instant to instant, so that there is produced 
in reality a revolving electrostatic field. Let the instantaneous 
displacements which issue from the three conductors per unit of 
axial length be denoted by q h q 2 , and q*, where the subscripts 1, 2, 
and 3 refer to the conductors A, B, and C respectively. To be 
consistent with the notation used before, these symbols should be 
provided with the " prime " sign, but this sign is omitted in order 
not to obscure the formulae. The displacements are considered 
positive when they are directed from the conductors into the die- 
lectric. Since electricity behaves like an incompressible fluid, as 
much of it as is displaced at any instant out of one conductor 
must be displaced into the other two conductors, so that at all 
times the following relation holds, namely, 

qi + q* + qz = (325) 

The three q's vary with the time according to the sine law. With 
a symmetrical spacing of the wires, and symmetrical voltages 
forming an equilateral triangle (Fig. 53), the effective values of 
the three q's are equal, and the corresponding instantaneous 
values are displaced in time phase by 120 degrees. The charg- 
ing current per unit length of a conductor is equal to the rate 
of change of the corresponding displacement with the time, or 

i = dq/dt (326) 

193 



194 THE ELECTRIC CIRCUIT [Art. 64 

But, with sinusoidal voltages, the displacements vary also ac- 
cording to the sine law, or 

q = Q m sm27rft, (327) 

where Q m is the maximum value of the displacement from one of 
the conductors. Substituting this value of q in eq. (326), we find 
that 

i = 2 7rfQ m cos 2 irft (328) 

Consequently, the amplitude of the charging current 

I m = 2TfQ m , . . . . . . . (329) 

and the same relation holds true for the effective values of the 
displacement and current. It is to be noted that the charging 
current leads the flux by 90 electrical degrees. Thus, knowing 
the displacement, the charging current can be calculated from 
eq. (329). If Q m is expressed in microcoulombs per kilometer, 
I m is in microamperes per kilometer. 

The actual charging current which flows through a cross- 
section of the conductor, is equal to that necessary to supply the 
displacement between this cross-section and the receiver end of 
the line. In other words, the charging current varies along the 
line, from a maximum at the generator end to zero at the receiver 
end. If the effective voltage along the line were constant in 
phase and magnitude, the amplitude of the charging current 
would vary according to a straight-line law. In reality, the volt- 
age varies along the line, due to its resistance and inductance, so 
that the variations in phase and amplitude of the charging cur- 
rent along the line follow a much more complicated law. 

The influence of the permittance of the line upon its voltage 
regulation is treated in Arts. 68 and 69 below. The problem here 
is a preliminary one; namely, with a given size and arrangement 
of conductors in a three-phase line, to find the permittance per 
kilometer of the equivalent single-phase line, for which the volt- 
age regulation is usually calculated. The problem is solved by 
applying again the principle of superposition. Each conductor 
is considered as forming a condenser with a .concentric cylinder 
of infinite radius, the three phases being star-connected and the 
three cylinders grounded. The vectors of the star and delta volt- 
ages are shown in Fig. 53, the subscripts 1, 2, 3 referring again 
to the conductors A, B, and C respectively. 



Chap. XVIIJ ELASTANCE OF THREE-PHASE LINES 



195 



Applying eq. (290) for the voltage between the conductors A 
and B, we have, for instantaneous values, 



e\2 = {(rqi/2ir) Ln (b/a) + (o-q 2 /2 t) Ln (a/b). 



(330) 



where, as before, the spacing is denoted by b, and the radii of the 
conductors by a. The first term on the right-hand side of this 
equation represents the action of system A, the second term that 
of system B. The action of the system C is equal to zero, be- 
cause, on applying eq. (290) for this system, it is observed that 
r = r' } on account of the symmetrical spacing. In other words, 




Fig. 53. Electric displacements in a three-phase line with symmetrical 
voltages and symmetrical spacing. 



for system C the conductors A and B lie on the same equipoten- 
tial cylindrical surface. The preceding equation is simplified to 
eu = (<2i — #2) S'j where S' is the elastance expressed by eq. 
(296), that is, the elastance between one of the conductors and 
the plane of symmetry 00' , as if the third conductor did not 
exist. Owing to symmetry, the other two equations are similar; 
thus we have 

ei2 = (31 - Q2)S f ; 

631 = (q-6 — Qi)S'. 



196 THE ELECTRIC CIRCUIT [Art. 64 

This result is interpreted graphically by Fig. 53, remember- 
ing that relations which hold true algebraically for instantane- 
ous values of sinusoidal quantities, hold true geometrically for 
the corresponding vectors of these quantities. According to eqs. 
(331), the instantaneous values of (qi — q 2 ), fe — qd, and (#3 — #1) 
are in phase with the corresponding voltages e i2 , e 2S , and e 3 i. For 
this reason, the vectors (Qi — Q 2 ), (Q 2 — Qs), and (Q 3 — Qi) are 
drawn in phase with the vectors E 12 , E 23 , and E Z i. In regard to the 
quantities Q h Q 2 , and Q 3 , we know that, for reasons of symmetry, 
they are equal numerically and are displaced in phase relatively 
to each other by 120 degrees. Therefore, they must be repre- 
sented by vectors from the center to the vertices of the triangle 
MNP. The condition is then fulfilled that each side of this tri- 
angle is equal to the difference of two vectors from the point 0. 

We see now that the three electric displacements Q h Q 2 , and 
Qz are in phase with the corresponding star- or Y-voltages of the 
system; also, from the similarity of the triangles, we have E12/E1 
= (Qi — Qzj/Qi, with corresponding relations for the other two 
phases. Consequently, eqs. (331) are reduced simply to 

E 1 = QiS'; 

E 2 = Q 2 S';\ (332) 

E3 = QsS f . 

We thus arrive at the following important conclusion: The dis- 
placement (and consequently the charging current) per phase of a 
three-phase line with symmetrical spacing and symmetrical voltages 
is equal to that in a single-phase line with the same conductors and 
the same spacing, provided that the star voltage of the three-phase 
line is equal to that between one conductor and the neutral plane 00' 
in the single-phase line. 

As explained in Art. 36, an equivalent single-phase line is 
obtained by taking one conductor of the three-phase line and 
assuming the transmission voltage to be equal to the star voltage 
of the actual transmission line; the return conductor is supposed 
to be devoid of both resistance and inductance. The preceding 
rule gives a simple method for finding the permittance of the 
equivalent line; namely, the permittance of the equivalent single- 
phase line is equal to that between one of the conductors of the actual 
line and the plane of symmetry between it and one of the other con- 
ductors, as if the third conductor did not exist. 



Chap. XVII] ELASTANCE OF THREE-PHASE LINES 197 

The calculation of the charging current with an unsymmetrical 
spacing of conductors is much more involved, and is explained in 
the next article. Fortunately, however, the spacing between the 
conductors affects the value of the charging current but little, 
with the usual ratios between size of conductor and spacing. 
The student can easily verify this fact by consulting any available 
table of capacities or charging currents of transmission lines. 
The reason for this is that the principal part of the elastance be- 
tween two small conductors occurs near the conductors, where 
the flux density is comparatively high. Consequently, it is pos- 
sible in practice to estimate the permittance per phase of a three- 
phase line with unsymmetrical spacing, by finding the limits of 
the permittance with symmetrical spacings. For instance, let 
two conductors be placed on a cross-arm and the third on top of 
the pole, forming an isosceles triangle. Let the spacings be 
2 m. and 1.6 m. respectively. The charging currents are differ- 
ent in the three conductors, but the average value is larger than 
with a symmetrical spacing of 2 m., and smaller than with a sym- 
metrical spacing of 1.6 m. Having found the charging currents 
or the equivalent permittances for these two spacings, one can 
assume an intermediate value by interpolation, or else take one of 
the two limits, whichever gives the more unfavorable operating 
conditions of the line. 

It is rather a tedious problem to estimate the influence of the 
ground upon the charging currents in a three-phase line. The 
theory is simple, the ground being replaced by the images of the 
three conductors, as in Fig. 51; but the formulae are long and in- 
volved, because the effects of six separate systems must be super- 
imposed. See problem 3 in the next article. 

Prob. 1. Show that when one of the conductors in a three-phase line 
fails, the charging current in the other two conductors drops to 86.6 per 
cent of its former value. Solution: Let C be the permittance between 
one of the conductors and the plane of symmetry between it and one of 
the other conductors. Then the charging current with the three phases 
alive is kC(E/v3), where E is the line voltage, and h is a coefficient 
of proportionality with which we are not concerned here. Operating 
single-phase, the charging current is hC{^E). The ratio of the two is 
0.5/ (1/ V3) = 0.866. 

Prob. 2. A three-phase, 140-kv., 25-cycle transmission line consists of 
conductors 2 cm. in diameter; the spacing is symmetrical and equal to 
3.5 m.; the length of the line is 250 km. What is the total reactive 



198 



THE ELECTRIC CIRCUIT 



[Art. 64 



power necessary to keep the line alive, and what are the voltage and the 
permittance per kilometer of the equivalent single-phase line? 

Ans. 7270 kva.; 80.8 kv.; 0.00947 mf. per km. 

Prob. 3. A three-phase transmission line consists of conductors 18 mm. 
in diameter, suspended all three in the same vertical plane, at a distance 
of 2.4 m. between the adjacent conductors. What are the limits of the 
elastance of the equivalent single-phase line? 

Ans. 100 and 113 megadarafs per km. 

Note: The proximity of the two limits shows that it is sufficient for 
practical purposes to consider the symmetrical spacing only, as far as 
the dielectric and magnetic effects are concerned. Mr. J. G. Pertsch, Jr., 
has called the author's attention to the fact that, with certain simplifying 
assumptions, and when the three wires are transposed, the equivalent 
spacing for inductance and capacity is equal to the geometric mean of 
the three actual spacings, or 

b eq = V 612623631. 

In the case under consideration the equivalent spacing is 3.02 m., and the 

corresponding elastance equals 105 mgd. per km. 

Prob. 4. Extend the treatment given in this article to the case in 

which the three delta voltages are different (Fig. 54), and show that the 

point coincides with the center 
of gravity of the triangle, a sym- 
metrical spacing of the conductors 
being presupposed as before. Solu- 
tion: Equations (331) hold true as 
before, and the sides of triangle 
MNP are parallel to those of 123, 
but the point cannot be deter- 
mined in this case from the sym- 
metry of the figure. Any point 
within the triangle 123 gives a set 
of star voltages E h E 2 , and E 3 , 
which will produce the given set of 
delta voltages; but there is only 

Fig. 54. Electric displacements in a one P oint ° from which the ravs 
three-phase line with unsymmetrical to the vertices of triangle MNP 
voltages and symmetrical spacing. satisfy condition (325). Since the 

displacements in the equivalent 
single-phase lines must be proportional to the voltages, condition (325) 
requires that the geometric sum of E h E i} and E 3 shall equal zero. 
The parallelogram 021 '3 gives the geometric sum of E 2 and E s equal 
to 01'. If is the correct point, 01' must be equal and opposite to 
01, and Om = §01. Similarly, the condition must be fulfilled that 
On = £02, and Op = £03. It is known from elementary geometry that 
the three bisectors of a triangle divide each other in the ratio of 2 to 1, 
and that the point of their intersection is the center of gravity of the 
triangle. Hence the point is the center of gravity of the triangle 123. 




Chap. XVII] ELASTANCE OF THREE-PHASE LINES 199 

From eqs. (331) we again derive eqs. (332), and finally arrive at the 
same conclusion as that printed in italics after these equations. The 
three star voltages being different, one from another, the three charging 
currents are also different, each leading the corresponding Q by 90 degrees. 
The permittance and the voltage of the equivalent single-phase line are 
also different for each phase, in spite of the symmetrical spacing of the 
conductors. 

Prob. 5. For a given three-phase line with symmetrical spacing and 
voltages, draw the electrostatic field for the instant when one of the 
delta voltages is equal to zero; also make three drawings of the field at 
the ends of intervals tV, tj and T 3 2 of a cycle later. Use the principle of 
superposition explained in Prob. 5, Art. 62, and apply it to the three 
component systems, A, B, and C, keeping in mind the relative magni- 
tudes of the instantaneous displacements. 

65. Three-phase Line with Unsymmetrical Spacing. 1 As 

is mentioned in the preceding article, the calculation of charging 
currents in a three-phase line with unsymmetrical spacing is 
much more involved than with symmetrical spacing, and is not 
of much practical importance at present. An outline of it is 
given here in order to fix more firmly in the student's mind the 
general principle of superposition, and the method by which the 
results are derived in the preceding article. Moreover, the in- 
fluence of the dielectric is becoming more and more important, as 
the transmission voltages and the lengths of transmission lines 
are increased. The time may come when the exposition given in 
this article will be of assistance in the solution of practical prob- 
lems. 

Let the three spacings be denoted by bi 2 , b 23 , and 6 3 i respec- 
tively. Equations (325) to (329) inclusive hold true as with a 
symmetrical spacing, but eq. (330) now becomes 

ei2 = (<rgi/2 tt) Ln (b 12 /a) + (<rq 2 /2 w) Ln (a/b 12 ) + 

(aq 3 /2w)Ln(b 23 /b 31 ), (333) 

because the effect of the system C is not equal to zero in this case. 
Similar equations may be written for e 23 and e 3 i, but only two 
equations are independent; the third is obtained by combining 
the two others, because the third voltage in a delta combination 
is determined by the other two voltages. The third independent 
equation is (325), and these three equations determine the three 
unknown g's. 

1 This article may be omitted, if so desired. 



(335) 



200 THE ELECTRIC CIRCUIT [Art. 65 

The following solution of these equations gives an insight into 
the physical relations, and leads to a result which is convenient 
in numerical work. The last term on the right-hand side of 
eq. (333) is usually much smaller than the other two terms, so 
that it may be conveniently represented in the form of a cor- 
rection to the other two, thus preserving the general form of eq. 
(330). Substituting the value of q 3 from eq. (325) into (333), we 
obtain 

e 12 = tai/2ir) Ln (b cl /a) - (aq 2 /2ir) Ln (b c2 /a), . (334) 

where the quantities, 

b c i = bi2hi/b 2 3, 
b c2 = b 22> b 12 /b n , 
b C 3 = bzib 2 3/b 12 , 

may be called the corrected spacings. The factors by which the 
displacements qi and q 2 are multiplied in eq. (334) are familiar, 
since they are of the same form as the right-hand member of 
eq. (296). It will be recalled that eq. (296) expresses the elas- 
tance between one conductor and the plane of symmetry of a 
single-phase line. The above-mentioned factors therefore rep- 
resent the elastances of single-phase lines having the corrected 
spacings b ci and b c2 respectively. Denoting the reciprocals of these 
elastances, or the corrected permittances, by C with the corre- 
sponding subscripts, eq. (334) and the two similar equations for 
the other phases are reduced to the form 

ei 2 = qi/Ci - q 2 /C 2 ; 
<?23 = q<ilC 2 — q 3 /C 3 ; 
esi = q 3 /C 3 — qi/Ci. 

On the other hand, for any star point 0, no matter where 
located, we have the following relation between the delta and 
star voltages: 

ei2 = ei — e 2 ; 
e 23 = e 2 — e 3 ; 
en = e 3 — ei. 



(336) 



(337) 



We again select the neutral point in such a manner that each 
star voltage is in phase with the corresponding q; that is, in 
phase quadrature with the corresponding charging current. Then 
the given three-phase system is directly resolved into three inde- 
pendent single-phase lines, and our problem is solved. If the 



Chap. XVII] ELASTANCE OF THREE-PHASE LINES 



201 



point is so selected, then by comparing eqs. (336) and (337) we 

have 

ei = qi/Ci;) 

e 2 = q 2 /cA (338) 

e 3 = qs/Cs. J 

Substituting the values of the g's from these equations into eq. 

(325) gives 

Ciei + C 2 e 2 + C 3 e 3 = (339) 

This is the condition which the point must satisfy if eqs. (338) 
are to hold true. Eliminating e 2 and e 3 from eq. (339) by means 
of the first and the last of the eqs. (337), and solving for e h we 
obtain 

d = e 12 C 2 /C - e 31 C 3 /C, (340) 

where 

C = Ci + C 2 + C 3 (341) 

As mentioned above, G, C 2 , and C 3 are the permittances per unit 
length between one wire and the plane of symmetry, for the cor- 
rected spacings defined by eqs. (335). 

Since relations which hold true algebraically for instantaneous 
values also hold true geometrically for the vectors of the same 
quantities, eq. (340) suggests a 
simple method for finding graph- 
ically the position of the neutral 
point in the vector diagram 
(Fig. 55). To locate 0, plot Ik 
= Ei 2 (C 2 /C) in the direction op- 
posite to #12, and W = E u {Cz/C) 
parallel to En. Or else, the prob- 
lem may be solved analytically, 
using either the orthogonal or 
the trigonometric form of com- 
plex quantities. Having deter- 
mined the position of 0, the 
three star voltages become known, 
and then the corresponding dis- 
placements are found from eqs. 
(338). The charging currents are determined by eq. (329), and 
are in leading quadrature with the corresponding star voltages. 
The given system is thus resolved into three independent equiva- 




Electric displacements in 



a three-phase system with unsym- 
metrical voltages and unsymmetrical 
spacing. 



(342) 



202 THE ELECTRIC CIRCUIT [Art. 65 

lent single-phase systems with the voltages E h E 2 , and E 3 , and 
the permittances d, C 2 , and Cz, per unit length. 

Instead of using the treatment given above, one could find the 
equivalent conductance and susceptance by using a method anal- 
ogous to that employed in Art. 63 of the Magnetic Circuit. 

Prob. 1. Determine the actual equivalent elastances in problem 3 of 
the preceding article, and compare them with the assumed limits. 

Prob. 2. Extend the treatment given above to the case in which the 
cross-sections of the three conductors are different, one from another. 

Prob. 3. Show how to estimate the influence of the ground upon the 
charging currents in a three-phase line, using the method of successive 
approximations. Solution: Replace the conducting ground by the three 
images A', B' ', and C of the actual conductors, as in Fig. 51. This gives 
six electric systems with cylinders at infinity. Applying the principle of 
superposition to the voltages between the conductors A — B and B — C, 
we get, by analogy with eq. (333) : 

<?i2 = (<^i/2 it) Ln (b n /a) + (aq</2w) Ln (a/b 12 ) + (crg 3 /27r) Ln (b 2S /b Z i) 

- (agi/2x) Ln (BA'/AA') - (<rq 2 /2ir) Ln (BB'/AB') 

- (crq 3 /2ir)Ln(BC'/AC'); 
e 23 = (o-g 2 /2 7r)Ln(6 23 /a)+ ((7# 3 /27r)Ln(a/& 23 ) + ((rgi/2 7r)Ln(& 31 /& 12 ) 

- (aq 2 /2^) Ln (CB'/BB') - (<rq s /2^) Ln (CC'/BC) 

- (aq^^LntfA'/BA'). 

From these two equations, together with eq. (325), the three unknown 
q's can be evaluated. By a method similar to that used in the text 
above, eqs. (342) are conveniently reduced to the form 

en + (<t/2tt) [gxLn {BA'/AA') + q 2 Ln (BB'/AB') + q 3 Ln (BC'/AC)] 1 

= gi/Ci - g 2 /C 2 ; Lr<ur> 

e 23 + ( ff /2x) [q 2 Ln(CB'/BB') +q 3 Ln (CC'/BC) + qi Ln(CA' /BA')] \ ^^ 
= q2/C 2 -q s /C 3 ; J 

where the three C"s and the corrected spacings are the same as before, 
without the ground. The last three terms on the left-hand side of eqs. 
(343) are small as compared to e n and e 23 , and represent the effect of the 
ground. Therefore, the simplest way of solving these equations is to 
neglect the correction terms in the first approximation, and to solve for 
the three q's exactly as explained in the text above, by finding the proper 
point (Fig. 55). The correction terms may be said to modify the 
values of e X2 and e 23 in eqs. (343). Having the values of the q's in the 
first approximation, the corrections are calculated, and, being added to 
e n and e 23 , give new values of the latter, say e'n and e' 23 . Having thus 
modified the triangle 123 in Fig. 55, a new point is found, and new 
values of the g's. The corrections for e J2 and e 2 3 can now be determined 
more accurately, and then new values of the q's found, which will be more 
nearly correct than the foregoing ones. In this way, the influence of the 
ground can be estimated with any desired degree of accuracy, without 
solving long and involved simultaneous equations. 



CHAPTER XVIII 

DIELECTRIC REACTANCE AND SUSCEPTANCE IN 
ALTERNATING-CURRENT CIRCUITS 

66. Dielectric Reactance and Susceptance. Let a condenser 
of permittance C, or elastance S, be connected across an alternat- 
ing-current line of voltage E and frequency /. Let any instanta- 
neous value of the voltage be denoted by e, where e = E m sin 2 Tft; 
then the corresponding instantaneous displacement in the dielec- 
tric is 

q = eC = e/S (344) 

This displacement varies according to the sine law and is in phase 
with the voltage, because q is at every instant proportional to e. 
The charging current flowing from the line into the condenser is 
at any instant equal to the rate of change of q with the time, or 

i = dq/dt = 2 wfCE m cos 2 irft (345) 

It will be seen from this equation that the charging current leads 
the voltage by 90 degrees, as has already been explained in Art. 48 
above. The amplitude of the charging current is 

I m = 2 wfCE m (346) 

It may thus be said that a permittance C connected across a 
source of voltage, of frequency /, is equivalent to a susceptance 

b=-I m /E m =-2 7rfC (347) 

The minus sign is necessary because the current is leading, while 
with a magnetic susceptance it is lagging. In other words, by 
using the minus sign in the case of dielectric susceptance, and the 
plus sign for magnetic susceptance, it is possible to extend the 
formulae deduced in Chapters 8 and 9 to alternating-current cir- 
cuits containing dielectrics. 

In the preceding formulae C is in farads, S in darafs, q in 
coulombs, and b in mhos. If C is expressed in microfarads and 
S in megadarafs, eq. (347) becomes 

b = - 2 tt/C X 10- 6 = - 2 tt/ X 10-VS mhos. . . (348) 
203 



204 THE ELECTRIC CIRCUIT [Art. 66 

The corresponding dielectric reactance is 

x = - 107(2 tt/C) = - 10 6 S/(2 wf) ohms. . . (349) 

The dielectric susceptance is equal to — 2 wfC only when there is 
no resistance in series with the condenser. When a condenser is 
connected in series or in parallel with an ohmic resistance, the 
treatment is analogous to that of a magnetic inductance in combi- 
nation with a resistance; that is, equivalent series and parallel com- 
binations are used, as explained in Art. 27. To give a detailed 
treatment here would simply be to repeat what has already been 
explained in the above-mentioned article. The only difference is 
that expressions (348) and (349) are used in place of (107) and 
(86), and the currents are leading, while with magnetic reactance 
they are lagging with respect to the impressed voltage. 

In some circuits both magnetic and dielectric susceptances 
are connected in parallel. They are simply added, taking into con- 
sideration their opposite signs. For instance, a magnetic suscept- 
ance of 7 mhos in parallel with a dielectric susceptance of 5 mhos 
is equivalent to a net magnetic susceptance of 2 mhos. A simi- 
lar rule is applied when magnetic and dielectric reactances are con- 
nected in series. 

With these explanations, the student will have no difficulty 
in dealing with any combination of resistances, condensers, and 
inductance coils in an alternating-current circuit. 

Prob. 1. A condenser of 7.3 mf. permittance is connected across a 
500- volt, 60-cycle supply. What are the susceptance and the charging 
current? 

Ans. b = —0.002754 mho; I = j 1.377 amp., the voltage being the 
reference vector. 

Prob. 2. The condenser in the preceding problem is shunted by a 
non-inductive resistance of 750 ohms. Find the total current and the 
power-factor. Solution: The current through the resistance is = 500/750 
= 0.6667 amp.; tan <*> = 1.377/0.6667 = 2.065; cos = 43.58 per cent 
(leading). Total current = 0.6667/0.4358 = 1.53 amp. 

Prob. 3. The condenser and the resistance in the preceding problem 
are connected in series, instead of in parallel. What is the equivalent 
parallel combination? 

Ans. C p - 1.387 mf.; r p = 926 ohms. 

Prob. 4. The voltage at the receiver end of a 25-cycle, single-phase 
transmission line is 45 + j 57 kv. ; the load current is 178 + j 69 amp. 
The series magnetic impedance of the line is 32 + j 68 ohms, and its 
capacity is 4.24 mf. Calculate the generator current and voltage. For 
purposes of calculation, one half of this capacity can be assumed to be 



Chap.XVHI] dielectric reactance 205 

connected across the generator end of the line, the other half across the 
receiver end. Solution: The dielectric susceptance at the receiver end 
of the line is -2tt X 25 X 2.12 X lO" 6 = -0.333 X lO" 3 mho. The 
corresponding charging current is 

j 0.333 X 10~ 3 (45000 + j 57000) = - 19 + j 15 amp. 

Consequently the total line current is 159 + j 84 amp. The line drop is 
(159 + j 84) (32 + j 68) = -624 + j 13500 volts. The generator volt- 
age is 44.38 + j 70.5 kv. The charging current at the generator end is 
j 0.333 (44.38 + j 70.5) = -23.5 + j 14.79 amp. The generator current 
is 135.5 + j 98.8 amp. 

Prob. 5. Explain the physical reason why a dielectric susceptance in- 
creases with the frequency, while a magnetic susceptance is inversely 
proportional to it. 

Prob. 6. Investigate the influence of a condenser in a circuit to which 
a non-sinusoidal voltage is applied; give a treatment similar to that in 
Art. 23. Show that the presence of an elastance accentuates higher 
harmonics in the current, while an inductance tends to diminish them. 
Make the physical reason for this difference clear to yourself. 

67. Current and Voltage Resonance. Let a condenser be 
connected in parallel with a pure reactance coil, across an alter- 
nating-current line. Let the current through the condenser be 
5 amp., leading, and that through the coil, 3 amp., lagging. 
Then the total current supplied from the generator is 2 amp., 
leading. Thus, we have the paradox that the resultant current is 
smaller than either of its components. It is even possible to 
adjust the permittance and inductance to such values as to make 
the leading and lagging components equal, in which case the gen- 
erator current is zero. This condition is called current resonance. 
When the line current is reduced to zero, total or perfect reson- 
ance takes place; otherwise the resonance is called partial. The 
condition for perfect resonance is that the lagging current shall be 
equal to the leading current, or, what "is the same, the dielectric 
susceptance must be numerically equal to the magnetic suscept- 
ance. Thus, if there is no resistance in either circuit, 

2tt/C = 1/(2x/L), 
from which 

27r/VCL = l (350) 

From this equation, any one of the three quantities /, C, and L 
can be determined, when the other two are given. Condition 
(350) may be fulfilled for the frequency of one of the higher har- 



206 THE ELECTRIC CIRCUIT ' [Art. 67 

monies of an e.m.f. wave, in which case we have partial resonance 
for the fundamental wave, and perfect resonance for one of the 
harmonics: If such is the case, the line current does not contain 
this harmonic, although it may be present to a considerable 
amount in the two branch currents. 

From the point of view of energy, current resonance consists 
in a periodic transformation of the potential energy of the elec- 
trostatic field into the kinetic energy of the magnetic field, and 
vice versa. When the current is at its maximum, the energy of 
the magnetic field of the reactance coil is also a maximum. But 
at this moment the voltage, and^ consequently the electrostatic 
displacement, are equal to zero, so that the whole energy of the 
circuit is in the magnetic field. One quarter of a cycle later, the 
displacement and the stored energy in the condenser are at a 
maximum, but the current and the magnetic field are equal to 
zero. At intermediate moments, the energy is contained partly 
in the electrostatic, and partly in the magnetic field. When 
condition (350) is satisfied, the maxima of the two energies are 
numerically equal, and the system " oscillates " freely in the 
electrical sense, in a manner analogous to the swinging of a pendu- 
lum. The generator merely maintains the necessary frequency, 
and supplies the i 2 r loss. Without this loss," it would not be 
necessary to have the generator at all; the oscillations, once 
started, would continue indefinitely at the proper frequency. 
When the two energies are not equal, there must be a cyclic ex- 
change of energy between the generator and one of the branches; 
namely, the one whose storage capacity for energy, at the gener- 
ator frequency, is larger than that of the other branch. We then 
have partial current resonance. 

The presence of resistance in either branch obscures the effect 
of resonance to some extent, leaving, however, its general char- 
acter unchanged. The best way to see the influence of resistance 
is to replace each impedance by its equivalent parallel combina- 
tion. We then have two pure susceptances with reactive currents, 
and two conductances, the currents through which are in phase 
with the line voltage. The energy supplied to the conductances 
is converted into heat, and thus does not enter into the electrical 
oscillations. 

Let now a dielectric reactance be connected in series with a 
magnetic reactance, across an alternating-current line. The cur- 



Chap. XVIII] DIELECTRIC REACTANCE 207 

rent through the two devices is the same, and may be taken as 
the reference vector. Let the dielectric reactance be such as to 
produce across the condenser a drop of 1000 volts, lagging behind 
the current by 90 electrical degrees. Let the voltage across the 
reactance coil be equal to 900 volts, leading the current by 90 
degrees. With these conditions, the total line voltage is equal to 
100 volts, lagging behind the current by 90 degrees. Thus, with 
a line voltage of only 100, it is possible to produce partial voltages 
of 1000 and 900 respectively. This condition is called voltage 
resonance. When the two reactances in series are equal, we have 
complete voltage resonance; otherwise the resonance is partial. 
The student will readily see that the condition for complete volt- 
age resonance is also expressed by eq. (350). In this case, the 
presence of resistance has no effect upon the correctness of the 
equation. By reading again the foregoing discussion of current 
resonance, and applying it to voltage resonance, the points of 
similarity and the differences between the two will be easily 
seen. 

One has to be on guard against possible resonance and a 
dangerous rise in potential in the operation of transmission lines 
and extended cable systems, because there the presence of per- 
mittance and inductance offers favorable conditions for surges 
between the dielectric and magnetic energies. These surges either 
produce large currents which open the circuit-protecting devices 
and interrupt the service, or the potential is raised to a value at 
which the insulation of the system is broken down. With a clear 
understanding of the principle of interchange of energy explained 
above, the student ought to be able to follow without difficulty 
special works on the subject. 1 

Prob. 1. A magnetic reactance of 65 ohms is connected in parallel 
with a permittance of 73.6 mf., across a 2200-volt, 25-cycle circuit. De- 
termine the total current, and the component currents, through the react- 
ance and through the condenser. 

Ans. 33.85 — 25.44 = 8.41 amp. (lagging). This is a case of partial 
current resonance, the total current being smaller than one of its com- 
ponents. 

1 See W. S. Franklin, Electric Waves; C. P. Steinmetz, Electric Dis- 
charges, Waves, and Impulses; also his larger work on Transient Electric 
Phenomena and Oscillations. Some elementary experiments and curves of 
resonance will be found in V. Karapetoff's Experimental Electrical Engineer- 
ing, Vol. II, Arts. 440 to 445. 



208 THE ELECTRIC CIRCUIT [Art. 68 

Prob. 2. The permittor and the reactance coil given in the preceding 
problem are connected across the same line in series, instead of in parallel. 
Find the total current and the component voltages. 

Ans. 102.3 amp. (leading); 2200 = 8850 - 6650 volts. This is a 
case of partial voltage resonance, the voltage drop across each of the 
two devices being larger than the applied voltage. 

Prob. 3. The elastance of a 60-cycle underground cable system is 
equal to 11.5 kilodarafs; at what value of the inductance in the circuit is 
resonance of the seventh harmonic to be feared? 

Ans. 1.65 millihenry. 

68. Voltage Regulation of a Transmission Line, Taking Its 
Distributed Permittance into Account. The voltage regulation 
of a transmission line, disregarding its permittance, is treated in 
Art. 33. The value of the permittance of a single-phase line is 
deduced in Art. 60, while in Chapter 17 the effect of the charging 
current in a three-phase line is considered, and it is shown how 
to calculate the permittance of the equivalent single-phase line. 
The inductance of transmission lines is treated in Chapter 11 of 
the author's Magnetic Circuit. It remains now to show how to 
determine, for a given load, the relation between the generator 
and receiver voltages of an equivalent single-phase line, knowing 
its constants; viz., the values of the distributed resistance, mag- 
netic reactance, and dielectric susceptance. 

Let the total resistance of the equivalent single-phase line be 
r ohms, and its magnetic reactance x ohms. Then the series 
impedance of the line is 

Z = r + jx (351) 

Let the dielectric susceptance of the line be b mhos, where b, ac- 
cording to eq. (347), is a negative quantity; and let the leakage 
conductance to the ground be g mhos. Then the shunted ad- 
mittance of the line is 

Y = g-jb (352) 

The leakage conductance is due to imperfect insulation of the 
line, and may also be made to take into account the corona loss, 
if any exists. The value of g can only be estimated, and in most 
cases may be safely neglected. It is introduced here in order to 
obtain a more general result, at the same time making the ex- 
pressions for Z and Y symmetrical. 

Since Y is uniformly distributed along the line, the current 
changes as the distance from the generator increases; and there- 



Chap. XVIII] DIELECTRIC REACTANCE 209 

fore it is necessary to consider the electrical relations in an in- 
finitesimal length ds, at some intermediate point of the line. Let 
the line voltage at this point be E, and the line current, /. The 
series impedance of the element ds is (Z/l) ds, and its shunted 
admittance is (Y/l) ds, where I is the total length of the line. 
Let dE be the increment in the voltage in the length ds, and let 
dl be the corresponding increment in the line current due to the 
shunted admittance. We have then 

dE =-I(Z/l)ds, (353) 

and 

dl = -E(Y/l)ds (354) 

The minus sign is needed on the right-hand side of eq. (353), 
because the drop in voltage / (Z/l) ds causes a decrement in E. 
Likewise the charging current E(Y/l)ds causes a decrement in 
the line current. 

Equations (353) and (354) contain two dependent variables, 
E and /. To eliminate /, we divide both sides of eq. (353) by ds 
and take the derivative with respect to s. The result is 

d*E/ds*=- (dI/ds)(Z/l). 

Substituting the value of dl /ds from eq. (354), we obtain 

d*E/ds* = EZY/P (355) 

This is a differential equation of the second order for E. We shall 
omit the solution of it and give only the result, for two reasons: 
first, because most students are not familiar with the methods of 
integration of differential equations; and secondly, because the 
solution is most conveniently expressed in hyperbolic functions of 
a complex variable, a form of function unknown to most students 
of engineering. 1 Fortunately, even for the longest transmission 

1 The simple theory of hyperbolic functions and the solution of eq. (355) 
may be found, among others, in the following works and articles: McMahon, 
Hyperbolic Functions; Dr. Kennelly, Applications of Hyperbolic Functions to 
Electrical Engineering; Dr. Steinmetz, Transient Electric Phenomena; Pender 
and Thomson, " The Mechanical and Electrical Characteristics of Transmission 
Lines," Trans. Amer. Inst. Electr. Engrs., Vol. 30 (1911); W. E. Miller, " Hyper- 
bolic Functions and Their Application to Transmission Line Problems," General 
Electric Review, Vol. 13 (1910), p. 177; M. W. Franklin, "Transmission Line 
Calculations," ibid., p. 74. For a proof of expansion (356), see Blondel and 
Le Roy, "Calcul des Lignes de Transport d'Energie a Courants Alternatifs en 
tenant compte de la Capacite et de la Perditance Reparties," La Lumiere 



210 THE ELECTRIC CIRCUIT [Art. 68 

lines built or projected, the solution can be represented with suffi- 
cient accuracy by a few terms of an infinite series, as follows: 

M x = E 2 (l + iYZ + J* Y 2 Z 2 + etc.) 

+ 7 2 Z(1 + i YZ + xk Y 2 Z 2 + etc.). (356) 

In this equation E x is the generator voltage, E 2 the receiver volt- 
age, the same as in Art. 33, and I 2 is the load current. Both Y 
and Z are known complex quantities, and therefore their product 
and the square of the product are also known. The terms in- 
volving Y 2 Z 2 are negligibly small in many cases. 

Instead of eliminating I from eqs. (353) and (354), E may be 
eliminated by a similar process, giving a differential equation for 
I, analogous to eq. (355). The solution of this equation is 

h = hiX + hYZ + -,\ Y 2 Z 2 + etc.) 

+ E 2 Y (1 + i YZ + T $* Y 2 Z 2 + etc.), (357) 

where I\ is the generator current, and I 2 the load current. 

The general form of eqs. (356) and (357) is the same as that of 
the corresponding equations in Art. 33, and in Chapters 11, 12, and 
13, so that the methods of calculation indicated there are appli- 
cable here, with self-evident modifications. 

Neglecting the leakage conductance g in eq. (352), and using 
the value of permittance given in Art. 60 above, also the value of 
inductance from Art. 61 of the Magnetic Circuit, we find that 

YZ = (Z/1000) 2 (-v.+ jw), . .. . . (358) ' 
and consequently 

Y 2 Z 2 = (Z/1000) 4 (v 2 - w 2 - 2jvw), . . . (359) 



where 
and 



v = 0.09514 (0.1 f) 2 [0.46 + 0.05/log (b/a)], . (360) 

w = 0.1515 ///log (b/a). . .... (361) 

In these expressions, I is the length of the line in kilometers, and 
r' is the resistance per kilometer of one conductor, in ohms. With 

Electrique, Vol. 7, 2nd Series (1909), p. 355; also J. F. H. Douglas, "Trans- 
mission Line Calculations," Electrical World, Vol. 55 (1910), p. 1066; and Dr. 
Steinmetz, Engineering Mathematics, p. 204. The best tables of hyperbolic 
functions are those published by the Smithsonian Institution; briefer tables 
will be found in McMahon's book and in the General Electric Review, Vol. 13, 
supplement to No. 5. See also Seaver's Mathematical Handbook, pp. 85 and 
266. 



Chap. XVIII] DIELECTRIC REACTANCE 211 

the extreme values of the ratio b/a of say 10 and 1000, the value 
of log (b/a) varies within the narrow limits of 1 to 3, so that the 
second term in the brackets in eq. (360) is comparatively small. 
In practice, the value of the whole expression in the brackets in 
formula (360) is usually between 0.48 and 0.50. This fact is 
taken advantage of in numerical calculations which do not require 
particular accuracy. 

Prob. 1. Check the numerical coefficients in formulae (360) and (361). 

Prob. 2. For a given receiver voltage, calculate the generator voltage, 
at no load and at full load, for some very long transmission line, the 
dimensions of which are taken from a descriptive article. 

Prob. 3. Solve problem 2 by the use of tables of hyperbolic functions, 
following the method indicated in one of the references in the footnote. 
Compare the results with those obtained in problem 2, and make clear to 
yourself the relative simplicity, and the limits of accuracy, of the series 
when one, two, or three terms are used. 

69. Approximate Formulae for the Voltage Regulation of a 
Transmission Line, Considering Its Permittance Concentrated at 
One or More Points. Instead of treating the permittance of a 
transmission line in the correct manner described in the preceding 
article, it is sometimes assumed to be concentrated at one or more 
points along the line. The calculation of voltage regulation then 
becomes similar to the treatment in Chapters 9 to 13. There is 
no particular advantage in this approximate treatment as far as 
the simplicity of numerical computations is concerned, because 
the formulae obtained are similar to eq. (356). It is advisable, 
however, for the student to deduce such formula? in order to see 
for himself that the form of eq. (356) is a rational one; more- 
over, this gives him one more exercise in the use of complex 
quantities. 

(a) The simplest assumption is to consider one half of the 
line permittance (and leakage, if any) concentrated at the gener- 
ator end of the line, the other half at the receiver end. The load 
current is in this case apparently increased by the current E 2 • J Y 
through the permittance \ Y connected in parallel with the load, 
so that the total receiver current is equal to I 2 + \ E 2 Y. Hence, 
the generator voltage is 

Ei = Z(I 2 + iE 2 Y)+E 2 , 
or 

Ei = E 2 (l + | YZ) + I 2 Z (362) 



212 THE ELECTRIC CIRCUIT [Art. 69 

Comparing this formula with eq. (356), we see that the principal 
terms are identical, the difference being in the additional terms 
containing higher powers of YZ. If the influence of the line per- 
mittance is small, for instance in short lines, the results calculated 
by means of both formulae differ from each other but very little. 
There is no reason, however, why the accurate expansion (356) 
should not be used in all cases, taking as many terms as are re- 
quired in a given problem. 

The generator current, with the capacity concentrated at both 
ends, is 

U = I 2 + i YE 2 + | YE h 

or, substituting the value of E\ from eq. (362), 

h = 1.(1 + i YZ) + E 2 Y(1 + i YZ). . . (363) 

This formula is similar to eq. (357), and differs from it only in 
the values of the coefficients of the minor terms. 

(b) The line permittance and leakage may also be concen- 
trated at the middle point of the line, in which case a diagram of 
connections is obtained similar to Fig. 42, except that the suscept- 
ance is dielectric and not magnetic. Introducing the voltage at 
the center of the line as an auxiliary unknown quantity, and 
eliminating it from the result, we obtain 

E, = E 2 (l + J YZ) + hZ(l + i YZ), . . (364) 
and 

h = |i(l + i YZ) + E 2 Y. . ... (365) 

(c) A closer approximation is obtained by assuming a part of 
the line permittance concentrated at the middle point, and the 
rest at both ends of the line. The fractions of the total permit- 
tance to be assigned to these three points are determined from 
Simpson's Rule for approximate integration; namely, according 
to this " parabolic " rule, 

y ave = [1/(3 n)] [y Q + 4 (2/1 + 2/3 + etc. +y n -i) 

+ 2 (2/2 + 2/4 + etc. + y n -2) + y n ], (366) 

where y aV e is the average ordinate of a given curve, n is the num- 
ber of equal parts into which the total width of the curve is sub- 
divided, and 2/0, 2/i> e ^c., are the actual ordinates at the points of 
division. In the above formula, n must be an even number. 
Let the given curve represent some arbitrary distribution of the 



Chap. XVIII] DIELECTRIC REACTANCE 213 

permittance along the line, and let n = 2. The foregoing formula 
gives 

C a „/ = MCo' + 4C7 + C 2 '), .... (367) 

where the C's are marked with the prime sign to indicate that 
they refer to unit length of the line. But in reality the permit- 
tance is uniformly distributed over the length of the line, so that 
C ave f = Co' = &' = C 2 '. Multiplying both sides of eq. (367) by 
the length I of the line, we obtain 

C = | C + | C + i C (368) 

This means that two thirds of the total permittance must be con- 
centrated at the middle of the line, and one sixth at each end. 1 

With this distribution of permittance it is again convenient 
to introduce the voltage at the center of the line as an auxiliary 
quantity. The relation between the load voltage and the gener- 
ator voltage is calculated in the well-known manner, by adding 
the voltage drop in the line to the load voltage. The result is 

#i = E 2 (l + hYZ+ & YW) + / 2 Z(1 + \ YZ) ; . (369) 
h = h(X + hYZ + & Y*Z*) + E 2 Y(1 + 5 % YZ 

+ v\*Y*Z*) (370) 

These formulae come closer to eqs. (356) and (357) than those 
obtained in the preceding two approximations. 

Prob. 1. Check formulae (364) and (365) by actually performing the 
algebraic transformations. 

Prob. 2. Check formulae (369) and (370) by actually performing the 
algebraic transformations. 

Prob. 3. If it be desired to have the permittance concentrated at 
five equidistant points along the line, show that according to Simpson's 
Rule one sixth of the total permittance must be placed in the middle, 
one twelfth at each end, and the rest at one quarter and three quarters 
of the length of the line. 

1 This result has been first indicated by Dr. Steinmetz, in his Alternating- 
Current Phenomena, in the chapter on " Distributed Capacity." 



APPENDIX, 



THE AMPERE-OHM SYSTEM OF UNITS. 

The ampere and the ohm can be now considered as two 
arbitrary fundamental units established by an international agree- 
ment. Their values can be reproduced to a fraction of a per 
cent according to detailed specifications adopted by practically 
all civilized nations. These two units, together with the centi- 
meter and the second, permit the determination of the values of 
all other electric and magnetic quantities. The units of mass and 
of temperature do not enter explicitly into the formula?, but are 
contained in the legal definition of the ampere and the ohm. 
The dimension of resistance can be expressed through those of 
power and current, according to the equation P = PR, but it is 
more convenient to consider the dimension of R as fundamental, 
in order to avoid the explicit use of the dimension of mass [M], 
Besides, there is no direct proof that the physical dimensions of 
electric power are the same as those of mechanical power. All 
we know is that the two kinds of power are equivalent one to 
the other. 

For the engineer there is no need of using the electrostatic or 
the electromagnetic units; for him there is but one ampere-ohm 
system, which is neither electrostatic nor electromagnetic. The 
ampere has not only a magnitude, but a physical dimension as 
well, — a dimension which, with our present knowledge, is fun- 
damental; that is, it cannot be reduced to a combination of the 
dimensions of length, time, and mass (or energy). Let the dimen- 
sion of current be denoted by [I] and that of resistance by [R]; 
let the dimensions of length and time be denoted respectively by 
the commonly recognized symbols [L] and [T]. The magnitudes 
and dimensions of all other electric units can be expressed through 
these four, as shown in the following table. For the correspond- 
ing expressions of the magnetic units in the ampere-ohm system, 
see Appendix I to the author's Magnetic Circuit. 

215 



216 



THE ELECTRIC CIRCUIT 



TABLE OF ELECTRIC UNITS, AND THEIR DIMENSIONS IN 
THE AMPERE-OHM SYSTEM 



Symbol and Formula 


Quantity 


Dimensions 


Name of the Unit 


J 


Current 


[I] 


Ampere 


U=I/A 


Current density 


[IL- 2 ] 


Ampere per square 
centimeter 


Q=IT 


Quantity of electricity 


[IT]* 


Coulomb (ampere- 




and dielectric flux 




second) 


D = Q/A 


Dielectric flux density 


[ITL-2] 


Coulomb per square 
centimeter 


E=IR 


Voltage, difference of 
potential, or e.m.f. 


[IR] 


Volt 


G=E/l 


Voltage gradient, elec- 
tric intensity, or di- 
electric stress 


[IRL- 1 ] 


Volt per centimeter 


r 


Resistance 


I [R] 




x = 2t/L 


Reactance 


Ohm 


z=(r 2 +x 2 )* 


Impedance 




g = l/r p 


Conductance 


) 




b = 1/xp 


Susceptance 


[ [R- 1 ] 


Mho 


y = l/z=(g*+b*)i 


Admittance 


) 




P = G/U 


Resistivity 


[RL] 


Ohm per centimeter 

cube 
Mho per centimeter 

cube 


y=U/G=l/p 


Conductivity 


[R-iL-i] 


S=E/Q 


Elastance 


[RT- 1 ] 


Daraf 


C = Q/E=l/S 


Permittance (capacity) 


[R-!T] 


Farad 


<r = G/D 


Elastivity 


[RT-!L] 


Daraf per centime- 
ter cube 


K =D/G=l/<r 


Permittivity 


[R-iTL-i] 


Farad per centime- 
ter cube 


P=EI 


Power 


[PR] 


Watt 


P' = P/V 


Density of power 


[PRL- 3 ] 


Watt per cubic cen- 
timeter 


W=hEQ 


Stored electric energy 


[PRT] 


Joule (watt-second) 


W'=hGD 


Density of electric en- 


[PRTL- 3 ] 


Joule per cubic cen- 




ergy 




timeter 


F = W/l 


Force 


[I2RTL- 1 ] 


Joulecen 


L = 2W/P 


Inductance 


[RT] 


Henry 



* These are also the dimensions of the electric pole strength. The con- 
cept of pole strength is of no use in electrical engineering, and, in the author's 
opinion, its usefulness in physics is more than doubtful. The whole elemen- 
tary theory of electrostatics can and ought to be built up on the idea of 
stresses and displacements in the dielectric, as is done in this work. 



THE AMPERE-OHM SYSTEM 217 

Other units of more convenient magnitude are easily created 
by multiplying the tabulated units by powers of 10, or by adding 
the prefixes milli-, micro-, kilo-, mega-, etc. 

A study of the physical dimensions of the electric and 
magnetic quantities is interesting in itself, and gives a better 
insight into the nature of these quantities. Moreover, for- 
mulae can be checked and errors detected by comparing phys- 
ical dimensions on both sides of the equation. Let, for instance, 
a formula for energy be given, 

W = aQDl/K, 

where a is a numerical coefficient. Substituting the physical di- 
mensions of all the quantities on the right-hand side of the equa- 
tion from the table below, the result will be found to be of the 
dimensions of energy. This fact adds to one's assurance that 
the given formula is theoretically correct. 

A slight irregularity in the system as outlined above is caused 
by the use of the kilogram as the unit force, because it leads to 
two units for energy and torque, viz., the kilogram-meter and 
the joule; 1 kg.-meter = 9.806 joules. Force ought to be measured 
in joules per centimeter length, to avoid the odd multiplier. Such 
a unit is equal to about 10.2 kg., and could be properly called 
the joulecen (= 10 7 dynes). There is not much prospect in sight 
of introducing this unit of force into practice, because the kilo- 
gram is too well established in common use. The next best 
thing to do is to derive formulae and perform calculations, when- 
ever convenient, in joulecens, and to convert the results into 
kilograms by multiplying them hy g = 9.806. 

Thus, leaving aside all historical precedents and justifica- 
tions, the whole system of electric and magnetic units is re- 
duced to this simple scheme: In addition to the centimeter, the 
gram, the second, and the degree Centigrade, two other funda- 
mental units are recognized, the ohm and the ampere. All 
other electric and magnetic units have dimensions and values 
which are connected with those of the fundamental six in a 
simple and almost self-evident manner, as shown in the table 
above. 

To appreciate fully the advantages of the practical ampere- 
ohm system over the C.G.S. electrostatic and electromagnetic 
systems, one has only to compare the dimensions, for instance, of 



218 



THE ELECTRIC CIRCUIT 



current density and voltage gradient in these three systems, as 
shown below. 



Dimension of current density . . 
Dimension of voltage gradient . 



The Am- 
pere-ohm 
System 



IIT 2 
IRL -1 



C.G.S. Electro- 
magnetic System 



L-^M'T-V" 4 
L*M*T-"V 



C.G.S. Electro- 
static System 






BIBLIOGRAPHY. 



Alternating-Current Phenomena, by Chas. P. Steinmetz. 

Theoretical Elements of Electrical Engineering, by Chas. P. Stein- 
metz. 

Alternating Currents, by Bedell and Crehore. 

Vectors and Vector Diagrams, by Cramp and Smith. 

Revolving Vectors, by G. W. Patterson. 

Electrical Engineering, by Thomalen. 

Die Wissenschaftlichen Grundlagen der Elektrotechnik, 
by Benischke. 

Die Wechselstromtechnik, by E. Arnold. 

Problems in Electrical Engineering, by Waldo V. Lyon. 

Electrical Problems, by Hooper and Wells. 

The Elements of Electrical Engineering, by Franklin and Esty. 

Principles of Electrical Engineering, by H. Pender. 

Modern Views of Electricity, by Oliver Lodge. 

Electric Waves, by W. S. Franklin. 

Elements of Electromagnetic Theory, by S. J. Barnett. 

Kapazitat und Induktivitat, by Ernst Orlich. 



219 



INDEX 

PAGE 

Addition and subtraction of sinusoidal currents and voltages 40 

Admittance, definition of 76 

expressed as a complex quantity or operator 89 

Admittances in parallel 76 

in series 80 

Air, dielectric strength of 165 

permittivity of 151 

Alternating-current power when current and voltage are in phase 45 

currents, advantages of 31 

currents, polyphase 99 

Ampere-ohm system of units 3, 215 

Amplitude factor, definition of 51 

Analogy, hydraulic, to an inductive circuit : . . . 65 

hydraulic, to Ohm's law 2 

hydraulic, to the dielectric circuit 145 

hydraulic, to the flow of electricity , 25 

mechanical, to a charged dielectric 159 

thermal, to Ohm's law 2 

thermal, to the flow of electricity 24 

Apparent power, definition of 56 

Average value of alternating current or voltage 50 

Cable, elastance of single-core 171 

insulation resistance of 26 

Cables, grading insulation of 174 

Capacity, see Elastance and Permittance. 

electrostatic, definition of 147 

specific inductive 151 

Charging current of condenser 203 

current of transmission line, see also Elastance 193 

currents of three-phase line with symmetrical spacing 196 

currents of three-phase line with unsymmetrical spacing 201 

Circle coefficient of induction motor 138 

diagram of induction motor or transformer * 136 

Circuit, alternating-current 31 

dielectric . ? 143 

dielectric, hydraulic analogue 145 

direct-current 1 

polyphase 99 

221 



222 INDEX 

PAGE 

Coefficient, leakage, of induction motor 138 

of self-induction see Inductance. 

temperature of electric resistivity 5 

Complex expression for admittance 89 

expression for impedance 88 

quantity, definition of 85 

Component, energy, of current or voltage 56 

reactive, of current or voltage 56 

Condenser, charging current of -203 

definition of , 143 

Conductance and resistance, how related, in an A.C. circuit 79 

definition of 2 

dielectric . 169 

Conductances, addition of 8 

Conductivity, definition of 14 

Conductor, definition of 1 

of variable cross-section 22 

unit, definition of 13 

Continuous current, see Current, direct. 

Core loss of transformer Ill 

Corona, electrostatic 167 

Current, alternating 31 

density, definition of 15 

direct 1 

due to non-sinusoidal voltage 71 

effective value of alternating 48 

energy and reactive components of 56 

primary, of induction motor 126 

radial flow of 26 

refraction, law of 28 

resonance • 205 

transient, in opening and closing a circuit. 71 

Currents, polyphase alternating 99 

Cycle of alternating wave, definition of 33 

Cylinders, elastance between two large parallel 188 

Daraf, definition of 148 

Delta-connected three-phase system 105 

Dielectric circuit 143 

conductance 169 

elastivity of 152 

energy stored in 158 

flux density 154 

flux, refraction of 164 

hysteresis 169 

nature of 144 

permittance of 147 



INDEX 



223 



PAGE 

Dielectric, permittivity of 151 

reactance 204 

strength 164 

stress 156 

susceptance 203 

Dimensions of units, table of 216 

Dispersion factor of induction motor 138 

Displacement, electric, illustrated 144 

Disruptive voltage, see Dielectric strength. 

Effective value in terms of harmonics 54 

value of variable current 49 

values of alternating currents and voltages, definition of 48 

Elastance between concentric spheres 175 

between small spheres 179 

between two large parallel cylinders 188 

definition of 148 

of a single-core cable 171 

of a single-phase line 176 

of a three-phase line with symmetrical spacing 196 

Elastances, addition of 149 

Elastivity, definition of „ 152 

Electric displacement illustrated . 144 

intensity, definition of 16 

intensity in the dielectric circuit 155 

power 10 

Electromotive force, see Voltage. 
Electrostatic, see also Dielectric. 

capacity, definition of 147 

corona 167 

field, nature of 143 

Energy component of current or voltage 56 

converted into heat 10 

density of 158 

stored in dielectric 158 

stored magnetic 62 

unit of electrical 11 

Equipotential surfaces defined 22 

Equivalent resistance, definition of 7 

series and parallel circuits 78 

sine-wave, definition of 53 

Exciting admittance of transformer Ill 

Exponential expressions for vectors and operators 97 

Farad, definition of 147 

Field, electrostatic, see Electrostatic field. 

Fleming's method for calculating the effective value of an irregular curve. 52 



224 INDEX 

PAGE 

Flux density, dielectric ...,..,..'..'. 154 

dielectric, see Dielectric flux. 

Form factor, definition of 51 

Fourier series r 43 

Frequency of alternating current or voltage, definition of 33 

Gradient, voltage, in the dielectric circuit 155 

voltage, definition of 16 

Ground, influence upon the charging currents in a three-phase line .... 202 

influence upon the elastance of a single-phase line 180 

Harmonics, definition of 41 

effects of elastance and inductance on . 205 

Heaviside, Oliver, nomenclature of 152 

Henry, definition of 62 

Heyland diagram of induction motor or transformer 136 

Homopolar machine 31 

Horse-power, English, defined 10 

metric, defined 10 

Hydraulic analogue of inductive circuit 65 

analogue of the dielectric circuit 145 

Hysteresis, dielectric 169 

Images, Kelvin's method of electric ... 180 

Impedance, definition of 67 

equivalent, of transformer 116 

expressed as a complex quantity or operator 88 

Impedances in parallel 80 

in series 68 

Inductance, definition of . . 60 

influence of, with non-sinusoidal voltage 69 

Induction motor, approximate analytical treatment 125 

characteristics with locked rotor 123 

circle coefficient or dispersion factor 138 

circle or Heyland diagram of 136 

equivalence to a polyphase transformer 122 

equivalent electrical diagram of 122 

exact analytical treatment 139 

input per phase 127 

magnetomotive forces in 124 

maximum output of 131 

primary current and power-factor 126 

pull-out torque of 130 

secondary resistance and reactance reduced to primary. 133 

slip, calculation of 126 

slip, defined 123 

squirrel-cage rotor 134 



INDEX 225 

PAGE 

Induction motor, starting torque of 129 

torque of 127 

Inductive reactance, see Reactance. 

Inertia as an analogue to inductance. 60 

Insulation, see also Dielectric. 

condenser type 175 

grading of 174 

Intensity, electric, definition of 16 

electric 155 

factor, illustrated 10 

Irregular paths, resistance and conductance of 27 

Joule, definition of 11 

relation of, to thermal units 11 

Joulecen, definition of 217 

Joule's law 10 

Kelvin's law of economy . 15 

Kelvin's method of electric images 180 

Kirchhoff's laws 17 

Law, Joule's 10 

Kirchhoff's first 18 

Kirchhoff's second 19 

of current refraction 28 

of flux refraction 163 

of economy, Kelvin's 15 

of minimum resistance 27, 160 

Ohm's, synopsis of 1 

Leakage conductance of transmission line 208 

factor of the induction motor 138 

Lehmann, Dr., method of finding resistance of irregular conductor 28 

Lehmann, Dr., method of mapping irregular field 162 

Line, see Transmission line. 

Magnetizing current of transformer Ill 

Mean, see Average. 

Mesh connection of polyphase system 101 

Mho, definition of 2 

Mi. um resistance, law of 160 

Motor, induction, see Induction motor. 

Neutral points of polyphase system 104 

Nomenclature xii 

Notation xiii 

Ohm's law, for an infinitesimal conductor 25 

hydraulic analogy to 2 



226 INDEX 

PAGE 

Ohm's law, synopsis of 1 

thermal analogy to 2 

Operator, admittance 89 

impedance 88 

Operators expressed as exponential functions . 97 

polar expressions for 93 

Output, maximum, of induction motor . . . . 131 

Parallel connection of conductors 7 

connection of impedances 80 

connection of susceptances and conductances 77 

Performance characteristics of the induction motor 122 

characteristics of the transformer 108 

characteristics of the transmission line 94, 208 

Permittance, see also Elastance. / 

definition of . . 147 

distributed, of transmission lines 208 

Permittances, addition of 149 

Permittivity, definition of 151 

relative 151 

Phase angle, definition of 34 

Phase displacement expressed by projections of vectors 91 

Polar coordinates, vectors and operators in 93 

Polyphase system, definition of 99 

system, neutral points of 104 

Power, alternating-current, when current and voltage are in phase 45 

apparent 56 

as double-frequency function 59 

average, of non-sinusoidal waves 57 

electric 10 

expressed by projections of vectors 91 

expression for average value in alternating-current circuit 56 

practical unit of 10 

real 56 

Power-factor, definition of 56 

of induction motor 126 

with non-sinusoidal waves 58 

Projections of vectors, addition and subtraction of 82 

Quadratic mean value 50 

Quantity factor, illustrated 10 

Quarter-phase system, star- and mesh-connected 101 

Radial flow of current 26 

Rayleigh, Lord, method of finding permittance of dielectrics of irregular 

shape 161 

method of finding resistance of irregular conductor 28 

Reactance and susceptance in an A.C. circuit 79 



INDEX 227 

PAGE 

Reactance, definition of 64 

dielectric 204 

equivalent, of transformer 118 

inductive 63 

leakage, of transformer 110 

secondary, of induction motor reduced to primary 133 

Reactive component of current or voltage 56 

Refraction of current 28 

of dielectric flux 164 

Regulation, see also Voltage regulation. 

speed, of induction motor 123 

voltage, of the transformer 108 

voltage, of the transmission line 94 

Resistance and conductance in A.C. circuits 79 

and temperature, relation between 5 

definition of 1 

equivalent, definition of 7 

equivalent, of transformer . : 118 

law of minimum 160 

secondary, of induction motor reduced to primary 133 

Resistances, addition of 8 

Resistivity, definition of 13 

Resonance, current 205 

voltage 207 

Series connection of admittances . 80 

connection of conductors 7 

connection of impedances 68 

Series-parallel combination of permittances and elastances 150 

combination of resistances 9 

Sine-wave, definition of 32 

definition of equivalent 53 

of current or voltage 31 

represented by a vector 36 

Single-phase line, effect of the ground upon the elastance of 180 

elastance of 176 

equations of lines of force and equipotential surfaces 184 

Sinusoidal currents and voltages, rule for addition and subtraction of . . . 40 

Slip, calculation of 126 

of induction motor defined 123 

Specific capacity, see Permittivity, relative, 
resistance, see Resistivity. 

Spheres, elastance between concentric 175 

elastance between small 179 

equations of lines of force and equipotential surfaces between . . 187 

Square root of mean square value, defined 49 

Star connection of polyphase system 101 



228 INDEX 

PAGE 

Steinmetz, Dr. C. P., symbolic notation of 83 

Stream lines, definition of 22 

Superposition, principle of 177 

Susceptance and reactance, how related, in an A.C. circuit 79 

definition of „ 75 

dielectric 203 

Susceptances in parallel 75 

Symbols, list of xiii 

System, four-wire, two-phase 99 

three-wire, two-phase 100 

polyphase, definition of 99 

quarter-phase, star- and mesh-connected 101 

three-phase, delta-connected 105 

three-phase, V- and T-connected 107 

three-phase, Y-connected 103 

T-connected three-phase system 107 

Temperature coefficient 5 

Thermal resistance, definition of 2 

Three-phase line, influence of the ground upon the charging currents: . . . 202 

with symmetrical spacing, elastance and charging current of . . . . 196 

with unsymmetrical spacing, charging currents of 201 

Three-phase system, delta-connected 105 

V- and T-connected 107 

Y-connected 103 

Time constant of electric circuit 72 

Torque of induction motor 127 

pull-out, of induction motor ; . . . 130 

starting, of induction motor 129 

Transformer, constant-potential, definition of 108 

core loss of Ill 

equivalent impedance of 116 

equivalent resistance and reactance of 118 

exciting admittance of Ill 

leakage reactance of 110 

magnetizing current of * . . Ill 

ohmic drop in , 110 

reactive drop in 110 

vector diagram of 113 

voltage ratio of 109 

voltage regulation 108, 115, 120 

Transient current in opening and closing a circuit 71 

Transmission line, see also Three-phase fine and Single-phase line. 

leakage conductance of 208 

. voltage regulation of 94 

voltage regulation, taking account of distributed per- 
mittance 208 



INDEX 229 

PAGE 

Tube of current, meaning of 23 

Two-phase, four-wire system 99 

three-wire system 100 

Unit conductor, definition of 13 

of conductance 2 

of electrical energy .. 11 

of resistance 1 

Units, C. G. S. and practical systems 3 

international electrical 3 

table of names and dimensions of 216 

the ampere-ohm system 215 

V-connected, three-phase system 107 

Vector, definition of 36 

diagrams, examples of 100, 102, 113 

used to represent a sine-wave 36 

Vectors, addition and subtraction of 37 

addition and subtraction of projections of 82 

expressed as exponential functions ' 97 

in polar coordinates 93 

Voltage, effective value of alternating 48 

energy and reactive components of 57 

gradient in the dielectric circuit 155 

gradient, definition of 16 

gradient, rupturing values of 165 

regulation of the transformer 108, 115, 120 

regulation of the transmission line 94 

regulation of transmission line, with permittance concentrated. 211 

regulation of transmission line, with permittance distributed. . . 208 

resonance 207 

Watt, definition of * 10 

Wave form of alternating current, or voltage 51 

representation, of irregular 41 

Y-connected, three-phase system 103 

Yrneh, definition of 148 



JUN 18 W2 



